Question

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 1

It is given that the sum of 7  terms and 17 terms is 49 and 289 respectively.

Let the first term of the AP be a and the common difference between the terms be d,

From the properties of arithmetic progression :

$S_n = \dfrac{n}{2}\{ 2a + ( n - 1 )d\}$. where $S_n$ is the sum of n terms, n is the number of terms, a is the first term and d is the common difference between the APs.

According to the condition : -

Case 1 : where the sum of 7 terms is 49

$\implies S_{7} =\dfrac{7}{2}\{2a+(7-1)d\}=49$

= > 49 x 2 = 7( 2a + 6d )

= > 7 x 7  x 2 = 7 x 2( a + 3d )

= > 7 = a + 3d

= > 7 - 3d = a           ...( i )

Case 2 : where the sum of 17 terms is 289

$\implies S_{17}=\dfrac{17}{2}\{2a+(17-1)d\}=289$

= > 289 x 2 = 17( 2a + 16d )

= > 17 x 17 x 2 = 17 x 2 ( a + 8d )

= > 17 = a + 8d

= > 17 - 8d = a        ...( ii )

Comparing the value of a from ( i ) and ( ii ) :

= > 7 - 3d = 17 - 8d

= > 8d - 3d = 17 - 7

= > 5d = 10

= > d = 2

Then, substituting the value of d in ( i )

= > 7 - 3d = a

= > 7 - 3( 2 ) = a

= > 7 - 6 = a

= > 1  = a

Hence,

Sum of n terms =$\dfrac{n}{2}\{2(1)+(n-1)2\}$

$S_n =\dfrac{n}{2}\{2+2n-2\}$

$S_{n}=\dfrac{n}{2}\times 2n$

$S_{n}=n^2$

Therefore the sum of n terms is n^2.

Answer 2

Refer to the attached Image

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