Question

find the common difference of ap whose first term is 5 and sum of the first four terms is half the sum of the next four terms ​

Answer 1

Answer:

a=5

S4=1/2

S4=n/2(2a+(n-1)d)

1/2=4/2(2×5+(4-1)d)

1=4(10+3d)

1=40+12da

-39=12d

-39/12=d

Answer 2

2 is the common difference of an AP .

Given:

a (first term of the arithmetic progression) = 5

$S_{4}=\frac{1}{2}(S_{8}-S_{4})$

To find:

d (Common Difference) = ?

Solution:

The general sequence of an AP is a ,a + d ,a + 2d ,a + 3d,…

Substituting a=5 then

5, 5 + d,5 + 2d,5 + 3d,5 + 4d,5 + 5d,5 + 6d,5 + 7d,,..

Let the first 4 terms be 5,5 + d,5 + 2d,5 + 3d

And let the next 4 terms be = 5 + 4d,5 + 5d,5 + 6d,5 + 7d

And $\bold{S_{4}=\frac{1}{2}(S_{8}-S_{4})}$----(1)

By substituting these values in (1)

$\begin{array}{l}{5+5+d+5+2 d+5+3 d} \\ {\qquad \qquad=\frac{1}{2}(5+4 d+5+5 d+5+6 d+5+7 d)}\end{array}$

20+6d=10+11d

10=5d

d=2

Therefore, the common difference = 2

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