Question

29. The sum of the first 7 terms of an AP is 49 and the sum of its first 17
terms is 289. Find the sum of its first n terms.​

Answer 1

Answer

• Sₙ = n²

Given

→  The sum of the first 7 terms of an AP is 49

→  The sum of its first 17  terms is 289

To Find

Sum of first n terms of AP

Formula

Sum of first n terms of an AP is denoted by ,

$\orange{\bigstar}\ \bf S_n=\dfrac{n}{2}(2a+(n-1)d)\ \; \green{\bigstar}$

Solution

Sum of first 7 terms of an AP is 49

$\to \rm S_7=\dfrac{7}{2}(2a+(7-1)d)\\\\\to \rm 49=\dfrac{7}{2}(2a+6d)\\\\\to \rm 49=7(a+3d)\\\\\to \rm a+3d=7...(1)$

Sum of first 17 terms of an AP is 289

$\to \rm S_{17}=\dfrac{17}{2}(2a+(17-1)d)\\\\\to \rm 289=\dfrac{17}{2}(2a+16d)\\\\\to \rm 17=a+8d\\\\\to \rm a+8d=17...(2)$

Subtract (1) from (2) ,

$\to \rm (a+8d)-(a+3d)=17-7\\\\\to \rm 5d=10\\\\\to \rm d=2$

On sub. d value in (1) , we get ,

$\to \rm a+3(2)=7\\\\\to \rm a=1$

Now , we need to find sum of it;s first n terms in this AP ,

$\to \rm S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\\to \rm S_n=\dfrac{n}{2}(2(1)+(n-1)(2))\\\\\to \rm S_n=\dfrac{n}{2}(2+2n-2)\\\\\to \rm S_n=\dfrac{n}{2}(2n)\\\\\to \rm S_n=n^2\ \; \pink{\bigstar}$

Answer 2

Answer

$\rm S_n=\dfrac{n}{2}(2a+(n-1)d)$

A/c to 1st condition ,

$\to \rm S_7=\dfrac{7}{2}(2a+(7-1)d)\\\\\to \rm 49=\dfrac{7}{2}(2a+6d)\\\\\to \rm 7=a+3d...(1)$

A/c to 2nd condition ,

$\to \rm S_{17}=\dfrac{17}{2}(2a+(17-1)d)\\\\\to \rm 289=\dfrac{17}{2}(2a+16d)\\\\\to \rm 17=a+8d...(2)$

Solve (2) - (1) ,

$\rm \to 10=5d\\\\\to \rm d=2$

Substitute " d = 2 " in (1) ,

$\to \rm a=1$

So ,

$\rm \to S_n=\dfrac{n}{2}(2(1)+(n-1)2)\\\\\to \rm \orange{\bigstar}\ \; S_n=n^2\ \; \green{\bigstar}$