24. Arithmetic Progression Solutions Part 1¶

1. First term of the sequence is $$t_1 = 1^2 + 4 = 5.$$ Second term of the sequence is $$t_2 = 2^2 + 4 = 8.$$ Similarly third term is $$t_3 = 3^2 + 4 = 13.$$ Difference between first two is $$8 - 5 = 3$$ and the difference between last two is $$13 - 8 = 5$$. Since the differences are not equal so the sequence is not an A.P.

2. $$t_3 = t_1 + t_2 = 3 + 5 = 8.$$

$$t_4 = t_2 + t_3 = 8 + 5 = 13.$$

$$t_5 = t_3 + t_4 = 13 + 8 = 21.$$

3. Clearly, common difference = $$d = 9 - 2 = 7$$. No. of terms = $$\frac{93 - 2}{7} + 1 = 14.$$

4. The first series is $$2, 4, 6, ..., 200$$ for $$100$$ terms and second series is $$3, 6, 9, ..., 240$$ for $$80$$ terms. Clearly, first common term is $$6$$ and common difference is also $$6$$ which is least common multiple of $$6$$. So common terms are $$6, 12, 18, ..., 198$$. So total no. of identical terms are $$\frac{198 - 6}{6} + 1 = 33.$$

1. $$t_1 = 4, t_2 = 7, t_3 = 10, ...$$

2. $t_1 = \frac{1}{2}, t_2 = \frac{2}{3}, t_3 = \frac{3}{4}, ...$
3. $t_1 = \frac{1}{2}, t_2 = \frac{4}{3}, t_4 = \frac{10}{3}, ...$
5. $$t_1 = 2a + b, t_2 = 4a + b.$$ Common difference = $$d = t_n - t-{n - 1} = 2a.$$ Since common term is independent of $$n$$ and is a constant therefore the sequence is in A.P.

6. $$t_7 = a + 6d \because c.d. = -1 - (-4) = 3 \text{ and } a = -4 t_7 = -4 + 18 = 14.$$

7. The first $$3$$ digit no. divisible by $$7$$ is $$105$$. Then we can find remaining by adding $$7$$ to them. So the sequence is $$105, 112, 119, 126, 133, 140, 147, 154, ..., 994$$.

8. Let first term is $$a$$ and common difference is $$d$$. Given $$7th$$ term is $$34$$ and $$12th$$ term is $$64$$. So $$a + 6d = 34$$ and $$a + 11d = 64.$$ So $$d = 6.$$ Therefore $$a = -2.$$ So 10th term is $$-2 + 9*6 = 52.$$

9. This problem is similar to 4 and is left as an exercise.

10. Let the terms are $$a - d, a, a + d.$$ Sum is $$15$$ so $$a = 5.$$

Sum of squares is $$83$$. Therefore $$3a^2 + 2d^2 = 83 \Rightarrow d = \pm 2.$$ So the terms are $$3, 5, 7$$ or $$7, 5, 3.$$

11. This problem is similar to 11 and has been left as an exercise to the reader.

12. This problem is similar to 11 and has been left as an exercise to the reader.

13. Given $$S_n = 5n^2 + 3n$$, thus, $$S_{n-1} = 5(n - 1)^2 + 3(n - 1)$$.

$$\therefore t_n = S_n - S_{n - 1} = 10n - 2$$

14. Given $$S_n = an^2 + bn$$, thus, $$S_{n-1} = a(n - 1)^2 + b(n - 1)$$

$$\therefore t_n = 2an - (a - b)$$

15. Clearly, $$t_1 = (a + b)^2$$ and $$c.d. = d = -2ab$$

$$\therefore S_n = n[a^2 + b^2 + ab(3 - n)]$$

16. Let $$n$$ be the number of sides of the polygon. From geometry the sum of angles of polygon is $$= (n - 2)180^{\circ}$$

Given $$a = 120^{\circ}$$ and $$d=5^{\circ}$$

$$\therefore$$ Sum on interior angles of the polygon

$$\Rightarrow \frac{n}{2}[120 + (n - 1)5]$$ (in degrees)

$$\Rightarrow \frac{n}{2}[240 + 5n - 5] = 180n - 360$$

$$\Rightarrow 5n^2 - 125n + 720 = 0$$

Roots of this equation are $$n = 9, 16$$

For $$n = 16$$ greatest angle is $$195^{\circ}$$ but from geometry no interior angle of a polygon can be greater than $$180^{\circ}$$, therefore no. of sides of the polygon is $$9$$.

17. The sum will be maximum till the first negative term. It is evident that $$a = 40$$ and $$d = -2$$.

$$t_n = 40 + (n - 1)\times -2 < 0$$

$$\therefore n > 21$$

Consequently, first 21 terms will give the maximum sum for this A. P.

$$S = \frac{21}{2}[80 + (21 - 1)\times -2] = 420$$

18. Given $$1 + 3 + 5 + 7 ... to n terms \ge 500$$

$$\therefore \frac{n}{2}[2 + (n - 1)2] \ge 500$$ or $$n^2 \ge 500$$

$$n \ge \pm \sqrt{500}$$ but $$n$$ is a positive number

$$n \ge \sqrt{500}$$ $$n \ge 22.36$$

Thus, least value of $$n$$ is 23.

19. Let $$a$$ be the first term and $$d$$ be the common difference of this A. P.

Given, $$mt_m = nt_n$$

$$\therefore m[a + (m - 1)d] = n[a + (n - 1)d]$$

$$a = d[1 - (m + n)]~~~~(\because m \neq n)$$

$$t_{m + n} = a + (m + n - 1)d = 0$$

20. Let $$x$$ by the first term and $$d$$ be the common difference of A. P.

Given, $$a = pth$$ term of A. P. $$= x + (p - 1)d$$

$$b = qth$$ term of A. P. $$= x + (q - 1)d$$

$$c = rth$$ term of A. P. $$= x + (r - 1)d$$

$$\therefore a - c = (p - r)d, b - a = (q - r)d, c - b = (r - q)d$$

$$\Rightarrow a(q - r) + b(r - p) + c(p - q)$$

$$\Rightarrow q(a - c) + r(b - a) + p(c - b)$$

$$\Rightarrow q(p - r)d + r(q - p)d + p(r - q)d = 0$$.

21. $$\because a, b, c$$ are in A. P.

$$\Rightarrow a - (a + b + c), b - (a + b + c), c - (a + b + c)$$ are in A. P.

$$\Rightarrow -(b + c), -(a + c), -(a + b)$$ are in A. P.

$$\Rightarrow b + c, a + c, a+b$$ are in A. P.

22. $$\because a^2, b^2, c^2$$ are in A. P.

$$\Leftrightarrow b^2 - a^2 = c^2 - b^2$$

$$\Leftrightarrow (b + a)(b - a) = (c + b)(c - b)$$

$$\Leftrightarrow \frac{b - a}{b + c} = \frac{c - b}{a + b}$$

$$\Leftrightarrow \frac{b + c - c -a}{(c + a)(b + c)} = \frac{c + a - a - b}{(a + b)(c + a)}$$

$$\Leftrightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b}- \frac{1}{c + a}$$

$$\therefore \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$$ are in A. P.

23. $$t_p = q = a + (p - 1)d$$ and $$t_q = p = a + (q - 1)d$$

Subtracting we get

$$q - p = (p - q)d \Rightarrow d = -1$$

Substituting the value of $$d$$ in any of the above terms’ equation

$$\therefore a = p + q - 1$$. Using these values of $$a$$ and $$d$$ to find the $$(p + q)$$ th term

$$t_{p + q} = a + (p + q - 1)d = 0$$

24. LHS = $$t_m + t_{2n + m}$$

$$\Rightarrow a + (m - 1)d + a + (2n + m - 1)d$$

$$\Rightarrow 2a + (2m + 2n -2)d$$

$$\Rightarrow 2[a + (m + n - 1)]d = 2t_{m + n}$$

25. $$\because \frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$$ are in A. P.

$$\Rightarrow \frac{a}{b + c} + 1, \frac{b}{c + a} + 1, \frac{c}{a + b} + 1$$ are in A. P.

$$\Rightarrow \frac{a + b + c}{b + c}, \frac{a + b + c}{c + a}, \frac{a + b + c}{a + b}$$ are in A. P.

Dividing by $$a + b + c$$

$$\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$$ are in A. P.

1. $$\because a, b, c$$ are in A. P.

Dividing by $$abc$$

$$\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$$ are in A. P.

2. $$\because a, b, c$$ are in A. P.

$$\Rightarrow b - a = c - b$$

$\Rightarrow \frac{1}{c - b} = \frac{1}{b - a}$

Multiplying both sides with $$ab + bc + ca$$

$\Rightarrow \frac{ab + c(b + a)}{c - b} = \frac{bc + a(b + c)}{b - a}$
$\Rightarrow ab(b - a) + c(b^2 - a^2) = bc(c - a) + a(c^2 - b^2)$

$$\Rightarrow b^2a + b^2c - a^2b -a^2c = c^2a + c^2b - b^2c - b^2a$$

$$\Rightarrow b^2(a + c) - a^2(b + c) = c^2(a + b) - b^2(c + a)$$

$$\therefore a^2(b + c), b^2(c + a), c^2(a + b)$$ are in A. P.

3. From previous part, we have

$$a^2(b + c), b^(c + a), c^(a + b)$$ are in A. P.

Dividing by $$abc$$ are in A. P.

$\Rightarrow \frac{ab + bc}{bc}, \frac{b(c + a)}{ca}, \frac{c(a + b)}{ab } \text{are in A. P.}$
1. $$\because (b - c)^2, (c - a)^2, (a - b)^2$$ are in A. P.

Adding $$a^2, + b^2 + c^2 -ab - bc - ca$$ to each of the terms and simplifying

$$(c - a)(a - b), (b - c)(a - b)$$ are in A. P.

Dividing each of the terms by $$(a - b)(b - c)(c - a)$$, we get

$$\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$$ are in A. P.

2. This problem is reverse of part ii of problem 27.

3. Let $$d$$ be the c. d. of A. P. Given, sum of first $$p$$ terms = 0

$\therefore \frac{p}{2}[2a + (p - 1)d] = 0$
$\because p \neq = 0 \therefore d = -\frac{2a}{p - 1}$

Sum of next $$q$$ terms = Sum of $$p + q$$ terms - Sum of $$p$$ terms

$= S_{p + q} - S_p$
\begin{align}\begin{aligned}= \frac{p + q}{2}\left[2a + (p + q - 1)d\right]~~~(\because S_p = 0)\\=\frac{p + q}{2}\left[2a + (p + q - 1)\left(-\frac{2a}{p - 1}\right)\right]\end{aligned}\end{align}
$- \frac{a(p + q)}{p - 1}.q$
4. Let the first term and common difference of the A. P. be $$a$$ and $$d$$ respectively.

Given, $$S_p = S_q$$

$\therefore \frac{p}{2}\left[2a + (p - 1)d\right] = \frac{q}{2}\left[2a + (q - 1)d\right]$
$2ap + p(p - 1)d = 2aq + q(q - 1)d$
$2a(p - q) + d(p^2 - q^2 -(p - q)) = 0$
$2a + (p + q - 1) = 0$
$S_{p + q} = \frac{p + q}{2}\left[2a + (p + q - 1)d\right] = 0$
5. Let the first term and common difference of the A. P. be $$a$$ and $$d$$ respectively.

Sum of latter half of $$2n$$ terms = $$S_{2n} - S_n$$

$= \frac{2n}{2}\left[2a + (2n - 1)d\right] - \frac{n}{2}\left[2a + (n - 1)d\right]$
$= \frac{n}{2}\left[4a + (4n - 2)d\right] - \frac{n}{2}\left[2a + (n - 1)d\right]$
$= \frac{n}{2}\left[2a + (3n - 1)d\right]$
$= \frac{1}{3}.\frac{3n}{2}\left[2a + (3n - 1)d\right]$
$= S_{3n}$
6. From given values

$S_1 = \frac{n}{2}\left[2.1 + (n - 1).1\right] = \frac{n(n + 1)}{2}.1$
$S_2 = \frac{n}{2}\left[2.2 + (n - 1).2\right] = \frac{n(n + 1)}{2}.2$
$S_3 = \frac{n}{2}\left[2.3 + (n - 1).3\right] = \frac{n(n + 1)}{2}.3$
$...$
$S_p = \frac{n}{2}\left[2.p + (n - 1).p\right] = \frac{n(n + 1)}{2}.p$

Adding all these we get

$S_1 + S_2 + ... + S_p = \frac{np}{4}(n + 1)(p + 1)$
7. Let $$x$$ be the first term and $$y$$ be the common difference of this A. P. Therefore,

$$a = \frac{p}{2}[x + (p - 1)y], b = \frac{q}{2}[x + (q - 1)y], c = \frac{r}{2}x + (r - 1)y]$$

$\therefore \frac{a}{p} = x + \frac{(p - 1)d}{2}$

Similarly,

$\frac{b}{q} = x + \frac{(q - 1)d}{2}$
$\frac{c}{r} = x + \frac{(r - 1)d}{2}$

Multiplying above three equations by $$(q - r), (r -p )$$ and by $$(p - q)$$ respectively and then adding gives us:

$\frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q)$
$= \left(x + \frac{(p - 1)d}{2}\right)(q - r) + \left(x + \frac{(q - 1)d}{2}\right)(r - p) + \left(x + \frac{(r - 1)d)}{2}\right)(p - q) = 0$
8. Let the first term of A. P be $$a$$ and c. d. be $$d$$.

Accordingly,

$S_m = \frac{1}{2}S_{m + n}$
$\therefore \frac{m}{2}\left[2a + (m - 1)d\right] = \frac{1}{2}\times\frac{m + n}{2}\left[2a + (m + n - 1)d\right]$

Let $$2a + (m - 1)d = x$$. Putting this in above equation

$2mx = (m + b)(x + nd)$
$2mx - mx - nx = (m + n)nd \Rightarrow x(m - n) = (m + n)nd$

Similarly,

$S_m = \frac{1}{2}S_{m + p}$
$\therefore x(m - p) = (m + p)nd$

Dividing these two we get

$\frac{m - n}{m - p} = \frac{(m + n)n}{(m + p)p}$
$(m + n)(m - p)n = (m + p)(m - n)p$

Dividing both sides by $$mnp$$ we hace

$(m + n)\left(\frac{1}{p} - \frac{1}{m}\right) = (m + p)\left(\frac{1}{n} - \frac{1}{m}\right)$
$\therefore (m + n)\left(\frac{1}{m} - \frac{1}{p}\right) = (m + p)\left(\frac{1}{m} - \frac{1}{n}\right)$
9. Let A. P. be $$a, a+d, a+ 2d, ..., a + nd$$

Sum of its odd terms

$S_{odd}= a + a + 2d + a + 3d + ... \text{to n + 1 terms}$
$S_{odd} = \frac{n + 1}{2}\left[2a + (n)2d\right]$
$\therefore S_{odd} = (n + 1)(a + nd)$

Sum of its even terms

$S_{even} = a + d + a + 3d + ... \text{to n terms}$
$S_{even} = \frac{n}{2}\left[2(a + d) + (n - 1)2d\right]$
$S_{even} = n(a + nd)$
$\therefore \frac{S_{odd}}{S_{even}} = \frac{n + 1}{n}$
10. Let $$a_1$$ and $$a_2$$ be the first terms and $$d_1$$ and $$d_2$$ be the common differences of the two series in A. P.

Given

$\frac{\frac{n}{2}\left[2a_1 + (n - 1)d_1\right]}{\frac{n}{2}\left[2a_2 + (n - 1)d_2\right]} = \frac{3n - 13}{5n + 21}$
(1)$\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{3n - 13}{5n + 21}$

Now,

$\frac{\text{24th term of first A. P.}}{\text{24th term of second A. P.}} = \frac{a_1 + (24 - 1)d_1}{a_2 + (24 - 1)d_2}$
$= \frac{a_1 + 23d_1}{a_2 + 23d_2} = \frac{2a_1 + 46d_1}{2a_2 + 46d_2}$

Putting $$n = 47$$ in (1), we get

$\frac{2a_1 + 46d_1}{2a_2 + 46d_2} = \frac{3\times 47 - 13}{5\times 47 + 21} = \frac{1}{2}$

Hence, the ratio is $$1:2$$.

11. Given $$t_m = \frac{1}{n}$$ and $$t_n = \frac{1}{m}$$

Thus,

\begin{align}\begin{aligned}a + (m - 1)d = \frac{1}{n}\\a + (n - 1)d = \frac{1}{m}\end{aligned}\end{align}

Subtracting $$d = \frac{1}{mn}$$. Substituting this in either $$t_m$$ or $$t_n$$, we get $$a = \frac{1}{mn}$$.

Thus,

$S_{mn} = \frac{mn}{2}\left[2a + (mn - 1)d\right]$

Substituting the values for $$a$$ and $$d$$, we get our desired result i.e.

$S_{mn} = \frac{mn + 1}{2}.$
12. Given, $$S_m = n, Sn = m$$. Therefore

\begin{align}\begin{aligned}\frac{m}{2}\left[2a + (m - 1)d\right] = n\\\frac{n}{2}\left[2a + (n - 1)d\right] = m\end{aligned}\end{align}

Simplifying, we get

$d = -\frac{2(m + n)}{mn}$

and

$a = \frac{m^2 + n^2 + mn - m - n}{mn}$

Now,

$S_{m + n} = \frac{m + n}{2}[2a + (m + n - 1)d]$

Substituting the values of $$a$$ and $$d$$, we get $$S_{m + n} = -(m + n)$$

13. It is given that

$S_1 = \frac{n}{2}\left[2a + (n - 1)d\right]$
$S_2 = \frac{2n}{2}\left[2a + (2n - 1)d\right]$

Subtracting, we get

$S_2 - S_1 = \frac{n}{2}\left[4a - 2a + (4n - 2 - n + 1)d\right]$
$S_2 - S_1 = \frac{n}{2}\left[2a + (3n - 1)d\right]$
$S_2 - S_1 = \frac{3n}{2.3}\left[2a + (3n - 1)d\right]$
$\Rightarrow S_3 = 3(S_2 - S_1).$
14. Let $$a$$ be the first term and $$d$$ be the common difference, then

$S = a + (a + d) + (a + 2d) + ... + (a + 2nd)$

and

$S_1 = a + (a + 2d) + (a + 4d) + ... + (a + 2nd)$

Consequently,

$S = \frac{2n + 1}{2}[2a + 2nd]$

and

$S_1 = \frac{n}{2}[2a + 2nd]$

Dividing we get the required result.

15. Let $$d$$ be the common difference for this A. P. Then from the given information

$b = a + 2d \Rightarrow d = \frac{b - a}{2}$
$c = a + (n - 1)d \Rightarrow n = \frac{2(c - a)}{b - a} + 1$

Adding $$a$$ and $$c$$, we get $$c + a = 2a + (n - 1)d$$

Let sum of $$n$$ terms be $$S$$, then

$S = \frac{n}{2}\left[2a + (n - 1)d\right]$
$\Rightarrow \left[\frac{c - a}{b - a} + 1\right][c -a]$

Consequently, we get the desired result.

16. Let $$a$$ be the first term and $$d$$ be the common difference of the given A. P. Then, from the given information

$t_p = a + (p - 1)d = x, ~~~ t_q = a + (q - 1)d$

Adding and subtracting these two we have

$x + y = 2a + (p + q - 2)d, ~~~ x - y = (p - q)d$

Now sum of $$p + q$$ terms is given by

$S = \frac{p + q}{2}\left[2a + (p + q - 1)d\right]$
$S = \frac{p + q}{2}\left[2a + (p + q - 2 + 1)d\right]$
$\therefore S = \frac{p + q}{2}\left[x + y + \frac{x - y}{p - q}\right]$
17. This problem is similar to 38 and has been left as an exercise.

18. Let $$a$$ be the first term and $$d$$ be the common difference of the given A. P. Then, from the given information

$\frac{S_m}{S_n} = \frac{m^2}{n^2}$
$\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}$
$\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$
$2a(n - m) + \{(m - 1)n - (n - 1)m\}d = 0$
$2a(n - m) + (m - n)d = 0 \Rightarrow a = \frac{d}{2}$
$\therefore a_m = a + (m - 1)d = \frac{d}{2} + (m - 1)d$
$a_m = \frac{2m - 1}{2}d, ~~~ a_n = \frac{2n - 1}{2}d$

Finally,

$\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}$
19. We can find a whole $$s$$ such that following equality takes place

$(2s + 1) + (2s + 3) + ... + (2s + 2n - 1) = n^k$
$(2s + n)n = n^k$
$s = \frac{2(n^{k - 2 - 1})}{2}$

But $$n$$ can be either even or odd. In both cases $$s$$ will be an integer, and our proposition is proved.

20. Let $$d$$ be the common difference for the A. P. Then, $$a_n = a_1 + (n - d)$$, but $$a_1 = 0 \therefore a_n = 0$$ $$\therefore a_2 = d$$

Consequently,

$S = \frac{2}{1} + \frac{3}{2} + ... + \frac{n - 1}{n - 2} - \left(1 + \frac{1}{2} + ... + \frac{1}{n - 3}\right)$
$= \sum_{k = 1}^{n - 2}\frac{k + 1}{k} - \sum_{k = 1}^{n - 2}\frac{1}{k} + \frac{1}{n - 2}$
$= \sum_{k = 1}^{n - 2}\left(1 + \frac{1}{k}\right) - \sum_{k = 1}^{n - 2}\frac{1}{k} + \frac{1}{n - 2}$
$= n - 2 + \frac{1}{n - 2} = \frac{(n - 2)d}{d} + \frac{d}{(n - 2)d}$
$= \frac{a_{n -1}}{a_2} + \frac{a_2}{a_{n - 1}}$
21. Multiplying both the numerator and denominator of each fraction by conjugate of the denominator, we get

$S = \frac{\sqrt{a_2} - \sqrt{a_1}}{a_2 - a_1} + \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3 - a_2} + ... + \frac{\sqrt{a_n} - \sqrt{a_{n - 1}}}{a_n - a_{n - 1}}$
$= \frac{1}{d}\left(\sqrt{a_2} - \sqrt{a_1} + \sqrt{a_3} - \sqrt{a_2} + ... + \sqrt{a_n} - \sqrt{a_{n - 1}}\right)$
$= \frac{\sqrt{a_n} - \sqrt{a_{n - 1}}}{d}$
$a_2 - a_1 = a_3 - a_2 = ... = a_n - a_{n - 1} = d$

Hence,

$S = \frac{\sqrt{a_n} - \sqrt{a_{n - 1}}}{d} = \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$
22. We have

$a_1^2 - a_2^2 = (a_1 - a_2)(a_1 + a_2) = -d(a_1 + a_2)$
$a_2^2 - a_3^2 = (a_2 - a_3)(a_2 + a_3) = -d(a_2 + a_3)$
$...$
$a_{2k - 1}^2 - a_{2k}^2 = (a_{2k - 1} - a_{2k})(a_{2k - 1} + a_{2k}) = -d(a_{2k - 1} + a_{2k})$
$S = -d(a_1 + a_2 + a_3 + ... + a_{2k - 1} + a_{2k}) = -d\frac{a_1 + a_{2k}}{2}2k$

But $$a_{2k} = a_1 + d(2k - 1), ~~~ a_1 - a_{2k} = -d(2k - 1)$$,

consequently

$S = -d(2k - 1)\frac{a_1 + a_{2k}}{2k - 1} = \frac{k}{2k - 1}(a_1^2 - a_{2k}^2)$