Question

What is the sum of the first 15 terms of an ap whose 11th and 7th term are 5.25 and 3.25 respectively

Answer 1

hope it helps mate
I think it is correct

Answer 2

The sum of the first 15 terms  is 56.25

Step-by-step explanation:

Formula of nth term : $a_n=a+(n-1)d$

Substitute n = 7

$a_7=a+(7-1)d$

$a_7=a+6d$

We are given that 7th term is 3.25

So,$a+6d=3.25$---1

Substitute n =11

$a_{11}=a+(11-1)d$

$a_{11}=a+10d$

We are given that 11th term is 5.25

So,$a+10d=5.25$

$a+6d+4d=5.25$

Using 1

$3.25+4d=5.25$

$4d=2$

$d=\frac{1}{2}=0.5$

Substitute the value in 1

a+6(0.5)=3.25

a=3.25-6(0.5)

a=0.25

Sum of first n terms : $S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 15

$S_{15}=\frac{15}{2}(2(0.25)+(15-1)0.5)$

$S_{15}=56.25$

Hence the sum of the first 15 terms  is 56.25

#Learn more:

If-15, -25 are respectively the 11th and 16 th terms of an AP, then find the sum of first 20 terms of the AP ?

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