Question

Find the sum of last 10 terms of the AP 8, 10, 12, ------126.​

Answer 1

Step-by-step explanation:

The sum of last ten terms of the AP: 8,10,12,...,126 is same as the sum of first ten terms of the AP: 126,124,122,...,8.

Therefore, a=126,d=−2 and n=10.

Now, S10=210[2×126+(10−1)×−2]

⇒S10=5×[252−18]=1170

Answer 2

Solution:-

Series:-

8 , 10 , 12 , ............... 126

Now find common difference

=> ( a₂ - a₁ ) => ( 10 - 8 ) = 2

Now we have to find last 10 term

Series:-

126 , 124 , 122 , 120 .............., 8

Now

a = 126 , d = - 2 and n = 10

Formula:-

Sn = n/2{2a + ( n - 1 )d}

Now

S10 = 10/2 { 2 × 126 + ( 10 - 1 ) × - 2 }

= 5 { 252 + (9)× - 2 }

= 5 { 252 - 18 }

=5{ 234 }

= 5 × 234

= 1170

Sum of last 10 term is 1170