Question

The sum of the 4th and the 8th of an AP is 24 and the sum of 6th and 10th is 44.Find the sum of the first five terms of the AP

Answer 1

Solution:-

Let:-

• The 1st term of AP=a
• The common difference=d

Given in 1st case:-

• The sum of 4th and 8th term of AP=24

Formula used:-

• An=a+(n-1)d

==>a+3d+a+7d=24

==>2a+10d=24

• dividing by 2 on both sides

==>a+5d=12-----(i)

Given in 2nd case:-

• The sum of 6th and 10th term of AP=44

==>a+5d+a+9d=44

==>2a+14d=44

• dividing by 2 on both sides

==>a+7d=22-----(ii)

• by solving equation 1 and 2 here

==>a+5d=12

==>a+7d=22

• by solving we get

==>-2d=-10

==>d=5

• putting the value of d=5 in eq 1

==>a+5d=12

==>a+25=12

==>a=12-25

==>a=-13

• Now, find the sum of 5th term of AP

Formula used here,

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

==>S5=5/2{2×(-13)+(5-1)×5

==>S5=5/2{-26+20}

==>S5=5/2×(-6)

==>S5=-15

Hence,

• The sum of 5th term of AP is -15

Answer 2

It is given that, the sum of the 4th and the 8th of an AP is 24 and the sum of 6th and 10th is 44.

We need to find the sum of the first five terms of the AP.

We know that,

$\bold{\displaystyle{\tt{a_n=a+(n-1)d}}}$

So,

$\displaystyle{\tt{a_4=a+3d}}\\\displaystyle{\tt{a_8=a + 7d}}\\\displaystyle{\tt{a_6=a+5d}}\\\displaystyle{\tt{a_{10}=a+9d}}$

A/q,

$\displaystyle{\tt{a_4+a_8=a+3d+a+7d=24}}$

$\displaystyle{\tt{\implies 2a+10d=24}}\\\displaystyle{\tt{\implies a+5d=12}}\\\displaystyle{\tt{\implies a=12-5d}}\:\:\:\:\:\:\:\: \cdots(i)$

and

$\displaystyle{\tt{a_6+a_{10}=a+5d+a+9d=44}}$

$\displaystyle{\tt{\implies 2a+14d=44}}\\\displaystyle{\tt{\implies a+7d=22}}\\\displaystyle{\tt{\implies 12-5d+7d=22}}\:\:\:\:\:\:\:\:\sf{[From\ (i)]}\\\displaystyle{\tt{\implies 12+2d=22}}\\\displaystyle{\tt{\implies d+6=11}}\\\displaystyle{\tt{\implies d=11-6}}\\\boxed{\displaystyle{\tt{\implies d=5}}}$

Putting d = 6 in eq.(i) :-

$\displaystyle{\tt{a=12-5d}}\\\displaystyle{\tt{\implies a=12-5(5)}}\\\displaystyle{\tt{\implies a=12-25}}\\\boxed{\displaystyle{\tt{\implies a=-13}}}$

Now, we know that, the sum of an AP is :-

$\displaystyle{\tt{S=\frac{n}{2}[2a+(n-1)d] }}$

In the AP,

• a = -13

• d = 5

As we have to find the sum of first 5 terms, so n = 5.

$\displaystyle{\tt{\implies S_5=\frac{5}{2}[2\times-13+(5-1)5] }}\\\\\displaystyle{\tt{\implies S_5=\frac{5}{2}[-26+5(4)] }}\\\\\displaystyle{\tt{\implies S_5=\frac{5}{2}(-26+20) }}\\\\\displaystyle{\tt{\implies S_5=\frac{5}{2}\times-6 }}\\\\\displaystyle{\tt{\implies S_5=5\times-3}}\\\\\boxed{\displaystyle{\tt{\implies S_5=-15}}}$

∴ So, the sum of the first 5 terms of the AP is -15.