Question

The sum of first 16terms of the AP: 10,6,2,..., is (a) ,-320,(b) 320,(c) -352,(d) -400​

Answer 1

a= 10 , d = 6-10= -4

sn = n/2[ a + an]

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

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

now , n= 16

so, s16 = 16/2[2*10 + (16-1)*(-4)]

=> s16 = 8[ 20 + 15(-4)]

=> s16 = 8[20 - 60]

=> s16 = 8[ - 40 ]

= -320

Answer 2

Answer:

Required sum of 16 terms is - 320.

Step-by-step explanation:

Given terms of an AS = 10 , 6 , 2 .....

Here,

First term of AP = a = 10

Common difference between the APs = d = 2 - 6 = 6 - 10 = - 4 .

From the properties of arithmetic progressions : -

• S${}_{n}=\dfrac{n}{2}$[2a+(n-1)d], where n is the number of terms, a is the first term, d is the common difference between the terms and S${}_n$ is the sum of n terms.

Here,

We have to find the sum of 16 terms, where first term is 10 and common difference is - 4.

Substituting values in the formula : -

= > S₁₆ = ( 16 / 2 )[ 2( 10 ) + ( 16 - 1 )( - 4 )  ]

= > S₁₆ = 8[ 20 + 15( - 4 ) ]

= > S₁₆ = 8[ 20 - 60 ]

= > S₁₆ = 8[ - 40 ]

= > S₁₆ = -320

Hence the required sum of 16 terms is - 320.