Sir please help me for solving the problem of class 10th math of arithmetic progressions of exercise 5.3 of math, how i solve this problem in easy way Find the sum of first 15 multiples of 8.
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The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
By the formula of sum of nth term, we know,
Sn = n/2 [2a+(n-1)d]
S15 = 15/2 [2(8) + (15-1)8]
= 15/2[6 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960
This is the video solution for question 13 exercise 5.3 chapter 5 Arithmetic Progression class 10 NCERT. Hope you will like it!!!