Question

Find the sum of n terms of an A.P whose n th term is given by tn = 5 - 6n.

Answer 1

given: tn = 5 - 6n
Put n=1, t1= -1
n=2, t2= -7
n=3,t3= -13

Sum of n terms= n/2[2a+(n-1)d]
= n/2[2(-1)+(n-1)-6]
= n/2[-2-6n+6]
=n/2(4-6n)
=> n(2-3n)

Answer 2

"The sum of n terms of an A.P is $S_{n}=2 n-3 n^{2}.$

An arithmetic progression (AP) is also called an arithmetic sequence, is a sequence of numbers which differ from each other by a common difference. For example, the sequence 3, 6, 9, 12,...  is an arithmetic sequence with the common difference 3.

Given:

$t_{n} = 5-6 n$

Answer:

We know that,

Sum of the n terms, $S_{n}=\frac{n}{2}(2 a+(n-l) d)$

$T_{1}=5-6=-1=a$

$T_{2}=5-(6 \times 2)=5-12=-7$

$T_{3}=5-(6 \times 3)=5-18=-13$

$d=t_{2}-t_{1}=-7+1=-6$

On substituting the values, we get,

$S_{n}=\frac{n}{2}(2(-1)+(n-1)-6)$

$=\frac{n}{2}(-2-6 n+6)$

$=\frac{n}{2}(4-6 n)$

=n(2-3 n)

$S_{n}=2 n-3 n^{2}$"