Question

Sum of first 5 terms of an A.P. is one fourth of
the sum of next five terms. If the first term=2,
then the common difference
of the A.P.is​

Answer 1

Answer:

let the A.P. be a,a+d,a+2d.......

We have T

1

+T

2

+T

3

+T

4

+T

5

=

4

1

[T

6

+T

7

+T

8

+T

9

+T

10

]

∵ sum =

2

n

(first term + last term)

Then

2

5

(T

1

+T

5

)=

4

1

×

2

5

(T

6

+T

10

)

⇒

2

5

(a+(a+4d))=

8

5

[(a+5d)(a+9d)]

⇒2a+4d=

4

1

[2a+14d]

⇒4a+8d=a+7d

⇒d=−3a ....(a=2)

⇒d=−3×2=−6

Answer 2

GIVEN :-

• Sum of first 5 terms of an AP is one fourth of the sum of next five

        terms.

• First term = 2

TO FIND :-

• Common difference of the AP .

SOLUTION :-

Sum = $\sf\dfrac{n}{2} \times \{First \ term + Last \ term \}$

According to the question ,

    $\sf\longrightarrow a_1+a_2+a_3+a_4+a_5 =\dfrac{1}{4} \times \left[a_6+a_7+a_8+a_9+a_{10} \right]$

   $\longrightarrow\sf \dfrac{5}{2}\times (a_1+a_5)=\dfrac{1}{4}\times \dfrac{5}{2} \times (a_6+a_{10}) \\\\\longrightarrow \dfrac{5}{2}\times (a+a+4d)=\dfrac{1}{4}\times \dfrac{5}{2}\times (a+5d+a+9d)\\\\\longrightarrow \dfrac{5}{2}\times (2a+4d)=\dfrac{1}{4}\times \dfrac{5}{2} \times (2a+14d) \\\\\longrightarrow 2a+4d= \dfrac{2a+14d}{4} \\\\\longrightarrow 4(2a+4d)=2a+14d \\\\\longrightarrow 8a+16d= 2a+14d \\\\\longrightarrow 8a-2a=14d-16d \\\\\longrightarrow 6a = -2d \\\\\longrightarrow\bf d= -3a$

 First term, a = 2

 d = -3 × 2

    = -6

Common difference of the AP = -6