Question

the first n terms of the progression. The first term and common difference of an arithmetic progression are a and -2. respectively. The sum of the first n terms is equal to the sum of the first 3n terms Express a in terms of n. Hence, show that n = 7 if a = 27.​

Answer 1

If a is the first term of an AP with common difference -2, then the first several terms are

a, a - 2, a - 4, a - 6, a - 8, …

with n-th term a - 2 (n - 1).

The sum of the first n terms is equal to the sum of the first 3n terms :

$\displaystyle \sum_{i=1}^n (a-2(i-1)) = \sum_{i=1}^{3n} (a-2(i-1))$

We have

$\displaystyle \sum_{i=1}^N (a-2(i-1)) = (a+2)\sum_{i=1}^N - 2\sum_{i=1}^Ni = (a+2)N - 2\cdot\dfrac{N(N+1)}2 = (a+1)N-N^2$

so that in the previous equation, the sums reduce to

$(a+1)n-n^2 = (a+1)(3n)-(3n)^2$

Solve for a :

$(a+1)n-n^2 = (a+1)(3n)-(3n)^2 \\\\ (a+1)n-n^2 = (a+1)(3n)-9n^2 \\\\ 9n^2-n^2 = (a+1)(3n)-(a+1)n \\\\ 8n^2 = (a+1)(2n) \\\\ 4n = a+1 \\\\ \boxed{a=4n-1}$

Now if a = 27, we have

27 = 4n - 1

28 = 4n

n = 7

as required.