Series Questions

Solution : Let a be the first term and r the common ratio of the given G.P. Then, $a_4$ = 54  and  $a_9$ = 13122 $\implies$  $ar^3$ = 54   and  $ar^8$  =  13122 $\implies$  $ar^8\over ar^3$ = $13122\over 54$  $\implies$  $r^5$ = 245  $\implies$  r = 3 Putting r = 3 in $ar^3$ = 54, […]

Solution : By using method of differences, The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP. Let $T_n$ be the nth term and $S_n$ denote the sum to n terms of the given series Then, $S_n$

Solution : By using method of differences, The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. Clearly,these differences are in AP. Let $T_n$ be the nth term and $S_n$ denote the sum to n terms of the given series Then, $S_n$

Solution : Let $S_n$ = 1 + 2 + 3 + ….. + n = $\sum_{k=1}^{n} k$ Clearly, it is an arithmetic series with first term a = 1, common difference d = 1 and last term l = n. $\therefore$ $S_n$ = $n\over 2$ (1 + n) = $n(n+1)\over 2$ Similar Questions Let $a_n$

Solution : Given, $a_2 + a_4 + a_6 + …… + a_{200}$ = $\alpha$      ………(i) and $a_1 + a_3 + a_5 + ….. + a_{199}$ = $\beta$           ………(ii) On subtracting equation (ii) from equation (i), we get ($a_2 – a_1$) + ($a_4 – a_3$) + ……… + (\(a_{200} –

Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 $\times$ 200 + $n\over 2$ { 2(240) + (n – 1) $\times$ 40 } = 11040 $\implies$  600 + $n\over 2$ {40(12+ n –

Solution : Let a be the first term and d (d $\ne$ 0) be the common difference of the given AP, then $T_{100}$ = a + (100 – 1)d = a + 99d $T_{50}$ = a + (50 – 1)d = a + 49d $T_{150}$ = a + (150 – 1)d = a + 149d

Solution : Since, x, y and z are in AP $\therefore$   2y = x + z Also,  $tan^{-1}x$, $tan^{-1}y$ and $tan^{-1}z$ are in AP $\therefore$   2$tan^{-1}y$ =  $tan^{-1}x$ +  $tan^{-1}z$ $\implies$ $tan^{-1}({2y\over {1 – y^2}})$ = $tan^{-1}({x + z\over {1 – xz}})$ $\implies$ $x + z\over {1 – y^2}$ = \(x + z\over {1 –

Solution : 0.7 + 0.77 + 0.777 + …… + upto 20 terms = $7\over 10$ + $77\over 10^2$ + $777\over 10^3$ +  ….. + upto 20 terms = 7[ $1\over 10$ +  $11\over 10^2$ + $111\over 10^3$ +  ….. + upto 20 terms ] = $7\over 9$[ $9\over 10$ +  $99\over 100$ + \(999\over

Solution : Let a, ar, $ar^2$ are in GP (r > 1) According to the question, a, 2ar, $ar^2$ in AP. $\implies$  4ar = a + $ar^2$ $\implies$ $r^2$ – 4r + 1 = 0 $\implies$ r = $2 \pm \sqrt{3}$ Hence, r = $2 + \sqrt{3}$    [ $\because$  AP is increasing] Similar Questions