Question

Find the sum of first 16th term of AP 10,6,2 ect
by using Sn=n\2 (2a+(n-nd)​

Answer 1

Answer:

a 2-a1

=10-6

=4

sn=n/2(2a+(n-1) d)

sn=16/2(2(10)+(16-1)4)

=8(20+15(4))

=8(20+60)

=8(80)

=640

Answer 2

Correct Question :

Find the sum of first 16th term of AP 10,6,2...By using Sₙ=n\2[(2a+(n-1)d].

Given :

• Sₙ = 16
• a = 10
• d = -4

To Find :

The sum of first 16th terms.

Solution :

Here we have to use the formula for finding the sum of n number of terms.

Required Formula :

$\boxed{\bf S_n=\dfrac{n}{2}[2a+(n-1)d]}$

where,

• Sₙ = Sum of respective term
• n = Respective Term
• a = First Term
• d = Common Difference

Explanation :

We know that if we are given the first term, common difference and the respective term and is asked to find the sum then our required formula is,

$\bf S_n=\dfrac{n}{2}[2a+(n-1)d]$

where,

• Sₙ = S₁₆
• a = 10
• d = 6 - 10 = -4
• n = 16

Using the required formula and substituting the required values,

$\\ :\implies\sf S_{16}=\dfrac{16}{2}[2(10)+(16-1)-4]$

$\\ :\implies\sf S_{16}=\cancel{\dfrac{16}{2}}[2\times10+(15)-4]$

$\\ :\implies\sf S_{16}=8[20+15\times(-4)]$

$\\ :\implies\sf S_{16}=8[20+(-60)]$

$\\ :\implies\sf S_{16}=8[20-60]$

$\\ :\implies\sf S_{16}=8[-40]$

$\\ :\implies\sf S_{16}=8\times-40$

$\\ :\implies\sf S_{16}=-320$

$\\ \therefore\boxed{\bf S_{16}=-320.}$

The sum of first 16th terms is -320.