If the 7th term of AP is 21,find the sum of first 13 terms of AP?

a7 = a + d(6) = 21

{so 2a+12 d = 42}

sum of first 13 = (13/2)(2a + 12d)

= (13/2)(42)

= 13 * 21

= 273

look at:
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html

To find the sum of the first 13 terms of an arithmetic progression (AP), we need to know the common difference (d) and the first term (a1).

In this case, we are given that the 7th term (a7) is 21. To find the common difference, we can use the formula for the nth term of an AP:

a_n = a1 + (n - 1) * d

Substituting n = 7 and a7 = 21 into the formula, we get:

21 = a1 + (7 - 1) * d
21 = a1 + 6d

Now, we have one equation with two variables (a1 and d). However, we can find another equation by noting that we have an arithmetic progression. The relationship between the terms of an AP is that the difference between any two consecutive terms is constant. Therefore, we can use the fact that a7 - a1 = 6d.

Substituting the values we know, we have:

21 - a1 = 6d

Now we have a system of two equations and two variables:

21 = a1 + 6d
21 - a1 = 6d

We can solve these equations simultaneously to find the values of a1 and d. Once we have the values, we can use the formula for the sum of the first n terms of an AP to find the sum of the first 13 terms:

S_n = n/2 * (2a1 + (n-1) * d)

Substituting n = 13, a1, and d into the formula will give us the sum of the first 13 terms of the AP.