Question

Find the sum first 15 terms of the A.P 8, 7 1/4, 6 1/2, 5 3/4
(step by step explanation)​

Answer 1

Answer:

=15/2(2×8+14×3/4)

=15/2((32+21) /2)

=15/2(53/2)

=795/4

=198.75 or 198 3/4

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Answer 2

Answer:

Sum of 15 terms of the A.P = 41$\frac{1}{4}$

Step-by-step explanation:

Given AP is

8, 7$\frac{1}{4}$, 6$\frac{1}{2}$, 5$\frac{3}{4}$ -----------

To find,

Sum of 15 terms of the A.P

Solution:

Given AP is 8, 7$\frac{1}{4}$, 6$\frac{1}{2}$, 5$\frac{3}{4}$ -----------

The first term of the AP =a =  8

The common difference = d = 7$\frac{1}{4}$ - 8 = $\frac{-3}{4}$

We know,

The sum to n terms of an AP = Sₙ= $\frac{n}{2}$[2a+(n-1)d], where 'a' is the first term and 'd' is the common difference of the AP

Substituting the value of a = 8, d = $\frac{-3}{4}$ and n = 15 we get

S₁₅= $\frac{15}{2}$[2×8+(15-1)× $\frac{-3}{4}$]

S₁₅= $\frac{15}{2}$[16+14× $\frac{-3}{4}$]

=  $\frac{15}{2}$[16+7× $\frac{-3}{2}$]

=  $\frac{15}{2}$× $\frac{11}{2}$

= $\frac{165}{4}$

= 41$\frac{1}{4}$

∴ Sum of 15 terms of the A.P = 41$\frac{1}{4}$

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