Question

The sum of 11 term of A.p is 891 .find the 30th term if the common different is 15

Answer 1

To find the 30th term of an arithmetic progression (AP) with a common difference of 15, we can use the formula:

term(n) = term(1) + (n-1) * d

Where term(n) is the nth term, term(1) is the first term, n is the number of terms, and d is the common difference.

Given that the sum of 11 terms is 891, we can use the formula for the sum of an arithmetic progression:

sum(n) = (n/2) * (2*term(1) + (n-1) * d)

Where sum(n) is the sum of the first n terms.

In this case, we can substitute n = 11 and sum(n) = 891 into the sum formula to solve for term(1):

891 = (11/2) * (2*term(1) + (11-1) * 15)
891 = 5.5 * (2*term(1) + 10 * 15)
891 = 5.5 * (2*term(1) + 150)
891 = 5.5 * (2*term(1) + 150)
891 = 11*term(1) + 825
891 - 825 = 11*term(1)
66 = 11*term(1)
66/11 = term(1)
term(1) = 6

Now that we know the first term is 6, we can use the term formula to find the 30th term:

term(30) = 6 + (30-1) * 15
term(30) = 6 + 29 * 15
term(30) = 6 + 435
term(30) = 441

Therefore, the 30th term of the arithmetic progression with a common difference of 15 is 441.