Question

can the sum of n terms in an AP is n square+13n​

Answer 1

Prove that "the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides". Find the value of y and Also verity:

Answer 2

yes , it is possible that sum of n terms of an AP is $n^2 +13n$ whose first term is 14 and common difference is 2

Step-by-step explanation:

using the formula of sum of n terms

i.e

$S_n = \frac{n}{2}(2a+(n-1)d)$..(1)

given that sum of n term is $n^2+13n$

then

$n^2+13n = n(n+13)\\\frac{n}{2}(2n+26)\\\frac{n}{2}(2n-2+28)\\\frac{n}{2}(2(n-1)+2\times 14)\\\frac{n}{2}(2(14)+(n-1)2)..(2)$

on comparing 1 and 2 we conclude that

a= 14 and d = 2

hence , it is possible that sum of n terms of an AP is $n^2 +13n$ whose first term is 14 and common difference is 2

#Learn more:

In an AP the sum of 1st n terms is 3n square /2+13n/2 find the 25th term

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