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If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer :

Given that,
$$S_7$$ = 49
$$S_{17}$$ = 289

We know, Sum of nth term;

$$\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]$$
$$\Rightarrow S_7= {{7} \over {2}} [2a + (n -1)d]$$
$$\Rightarrow S_7 = {{7} \over {2}} [2a + (7 -1)d]$$
$$\Rightarrow 49 ={{7} \over {2}} [2a + 6d]$$
$$\Rightarrow$$ 7 = (a+3d)
$$\Rightarrow$$ a + 3d = 7 …........ (i)

In the same way,

$$\Rightarrow S_{17} = {{17} \over {2}} [2a + (17-1)d]$$
$$\Rightarrow 289 = {{17} \over {2}} (2a + 16d)$$
$$\Rightarrow$$ 17 = (a + 8d)
$$\Rightarrow$$ a + 8d = 17 ……… (ii)

Subtracting equation (i) from equation (ii),
5d = 10
d = 2

From equation (i), we can write it as;
$$\Rightarrow$$ a + 3(2) = 7
$$\Rightarrow$$ a + 6 = 7
$$\Rightarrow$$ a = 1

Hence,
$$\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]$$
= $${{n} \over {2}}[2(1)+(n – 1)×2]$$ = $${{n} \over {2}}(2+2n-2)$$ = $${{n} \over {2}}(2n)$$ = $$n^2$$