Question

The sum of the first and 8 terms and first 24 terms of an A.P are equal. Find
the sum of its 32 terms.

please answer it fast it's urgent​

Answer 1

Answer:

0

Step-by-step explanation:

According to the question, Sum of the first 8 terms of an AP is equal to the Sum of the first 24 terms of the same AP.

We know that the Sum of n terms can be written as:

$\Longrightarrow \rm S_{n} = \dfrac{n}{2} \bigg[2a + (n-1)d\bigg]$

Where:

• n = Number of terms.
• a = First term.
• d = Common difference.

Using the above relation and from the statement given in the question we can say that:

$\Longrightarrow \rm S_{8} = S_{24}$

$\Longrightarrow \rm \dfrac{n}{2} \bigg[2a + (n-1)d\bigg] = \dfrac{n'}{2} \bigg[2a + (n'-1)d\bigg]$

[Taking n as n' for S₂₄ as the number of terms are different]

$\Longrightarrow \rm \dfrac{8}{2} \bigg[2a + (8-1)d\bigg] = \dfrac{24}{2} \bigg[2a + (24-1)d\bigg]$

$\Longrightarrow \rm 4 \bigg[2a + 7d\bigg] = 12 \bigg[2a + 23d\bigg]$

Cross multiply:

$\Longrightarrow \rm \dfrac{2a + 7d}{2a + 23d} = \dfrac{12}{4}$

$\Longrightarrow \rm \dfrac{2a + 7d}{2a + 23d} = \dfrac{3}{1}$

Cross multiply again:

$\Longrightarrow \rm 2a + 7d = 3(2a + 23d)$

$\Longrightarrow \rm 2a + 7d = 6a + 69d$

$\Longrightarrow \rm 2a - 6a = 69d - 7d$

$\Longrightarrow \rm -4a = 62d$

$\Longrightarrow \rm \dfrac{-4a}{62} = d$

$\Longrightarrow \boxed{\rm d = \dfrac{-2a}{31}}$

Now that we have the value of the common difference (d) , let's find the sum of the first 32 terms.

$\Longrightarrow \rm S_{n} = \dfrac{n}{2} \bigg[2a + (n-1)d\bigg]$

$\Longrightarrow \rm S_{32} = \dfrac{32}{2} \bigg[2a + (32-1)d\bigg]$

$\Longrightarrow \rm S_{32} = 16\bigg[2a + 31d\bigg]$

Substitute the value of 'd' [common difference] here.

$\Longrightarrow \rm S_{32} = 16\bigg[2a + 31\bigg[\dfrac{-2a}{31}\bigg]\bigg]$

$\Longrightarrow \rm S_{32} = 16\bigg[2a - 2a\bigg]$

$\Longrightarrow \rm S_{32} = 16\bigg[0\bigg]$

$\Longrightarrow \rm S_{32} = 0$

Therefore the sum of the first 32 terms of the AP is 0.

Answer 2

given :- sum of 4

th

and 8

th

term is 24.

a+3d+a+7d=24

2a+10d=24

a+5d=12.....(i)

sum of 6

th

and 10

th

term is 44.

a+5d+a+9d=44

12+a+9d=44......from (i)

a+9d=32

a+5d+4d=32

12+4d=32

d=5

a+25=12

a=−13

AP=−13,−8,−3,2,7,12....