Question

(i) If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

(ii) If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Answer 1

(i)

In the given problem, we need to find the sum of n terms of an A.P. Let us take the first term as a and the common difference as d.

Here, we are given that,

So, as we know the formula for the sum of n terms of an A.P. is given by,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 7, we get,

Further simplifying for a, we get,

Also, using the formula for n = 17, we get,

Further simplifying for a, we get,

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (3),

Now, using the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of first n terms for the given A.P. is.

(ii) Given:
$$\begin{gathered} S_4=40 \\ {S}_{14}=280 \\ \mathrm{Now}, \\ {S}_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ {S}_4=\frac{4}{2}\left[2a+\left(4-1\right)d\right] \\ \Rightarrow 40=2\left[2a+3d\right] \\ \Rightarrow 2a+3d=20...\left(1\right) \\ {S}_{14}=\frac{14}{2}\left[2a+\left(14-1\right)d\right] \\ \Rightarrow 280=7\left[2a+13d\right] \\ \Rightarrow 2a+13d=40...\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 10d=20 \\ \Rightarrow d=2 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ a=7 \\ \mathrm{Therefore},a=7\mathrm{and}d=2 \\ \mathrm{Thus}, \\ {S}_n=\frac{n}{2}\left(2\times 7+\left(n-1\right)2\right) \\ =\frac{n}{2}\left(14+2n-2\right) \\ =\frac{n}{2}\left(12+2n\right) \\ =n\left(6+n\right) \\ =n^2+6n \end{gathered}$$

Hence, the sum of its first n terms is n² + 6n.