# Series Questions

## If the 4th and 9th terms of a G.P. be 54 and 13122 respectively, find the G.P.

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, …

## Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….

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, $$S_n$$ …

## Find the sum to n terms of the series : 3 + 15 + 35 + 63 + …..

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, $$S_n$$ …

## Prove that the sum of first n natural numbers is $$n(n+1)\over 2$$

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 $$a_n$$ …

## If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

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 …
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