Question

the nth term of an AP is given by Tn=10-6n. Find the sum of first n terms of the AP

Answer 1

Let nth term be the 1st term of the given arithmetic progression.

Given, T${}_{n}$= 10 - 6n

∴ T₁ = 10 - 6( 1 )

  T₁ = 10 - 6

  T₁ = 4

We know that the sum of n terms of any arithmetic progression is $\dfrac{n}{2} [ T_{1} + T_{n}]$

∴$S_{n} = \dfrac{n}{2}[ 4 + 10 - 6n ]$

⇒ $S_{n} = \dfrac{n}{2}[ 14 - 6n]$

⇒ $S_{n} = \dfrac{n}{2} \times 2( 7 - 3n )$

⇒ $S_{n} = n( 7 - 3n )$

⇒ $S_{n} = 7n - 3n^2$

Therefore the sum of n terms of the given AP is 7n - 3n^2.

Answer 2

last term = 10-6n

at n = 1
= 10-6(1)
= 4
so, a= 4
at n= 2
= -2
a+d = -2

so, d = -6

Sn = n/2(2a+(n-1)-6)
=. n/2(14+6n)
= 7n+3n² ans..

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