Question

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Answer 1

Given that the sum of first 9 terms of an AP is 81.

We know that sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

= > (9/2)[2a + (9 - 1) * d] = 81

= > (9/2)[2a + 8d] = 81

= > 9(2a + 8d) = 162

= > 2a + 8d = 18

= > a + 4d = 9  ------ (1)


Given that sum of its first 20 terms is 400.

= > (20/2)[2a + (20 - 1) * d] = 400

= > 10(2a + 19d) = 400

= > 2a + 19d = 40        ------------- (2)


On solving (1) * 2 & (2), we get

= > 2a + 8d = 18

= > 2a + 19d = 40

      ----------------------

              -11d = -22

                   d = 2



Substitute d = 2 in (1), we get

= > a + 4d = 9

= > a + 4(2) = 9

= > a + 8 = 9

= > a = 1.

Therefore, The first term is 1 and the common difference d = 2.


Sum upto 15th term s15 = (n/2)[2a + (n - 1) * d]

= > (15/2)[2(1) + (15 - 1) * 2]

= > (15/2)[2 + 14 * 2]

= > (15/2)[30]

= > 15 * 15

= > 225.



Hope this helps!

Answer 2

Put D = 2 in equation (3)


A = ( 9 - 4D )


A = ( 9 - 4 × 2 ) = 9-8 = 1


First term ( A ) = 1

And, Common Difference ( D ) = 2



Sn = N/2 × [ 2A + ( N - 1 ) × D ]



S15 = 15/2 × [ 2 × 1 + ( 15-1) × 2 ]



=> 15/2 × ( 2 + 28 )


=> 15 /2 × 30



=> ( 15 × 15 ) = 225



Hence,


Sum of first 15th term = 225.



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