Question

find the sum of the terms of AP whose first term, last term and common difference are 3,101 and 7 respectively

Answer 1

Answer:

n=(xn-x1)÷d+1

=15

Sn=n÷2(x1+xn)

=15÷2(3+101)

=780

Answer 2

Sum of the terms in AP is 780

Given:

First term a = 3  and  Last term a$_{n}$ = 101  

Common difference d = 7

To find:

Sum of the terms in AP from given data  

Solution:

To find the sum of the terms we need to know number of term, find number of terms in given AP as given below

as we know nth term of AP = a + (n-1) d

From data 3 + (n-1)7 = 101

⇒ (n - 1)7 = 98

⇒ n - 1 = 14

⇒ n = 15

Number of term in AP, n = 15

Sum of 15 terms in AP = $\frac{n}{2} [a_{1} + a_{n} ]$  

⇒ $\frac{15}{2} [3+101 ]$ = $\frac{15}{2} [104 ]$ = 15 [52] = 780

Sum terms in AP = 780

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