Question

Find the common difference of an AP whose first term is 5
and the sum of the first four terms is half the sum of the next
four terms.​

Answer 1

Answer:

Answer:

Given :-

First Term = 5

Sum of first 4 terms = half of next two terms

To Find :-

Common Difference

Solution :-

We know that

\large \sf \: Sum \: = \dfrac{n}{2} \bigg(a + l \bigg) \:Sum=

2

n

(a+l)

Now,

Sum = 4/2 (5 + 5 + 3d)

Sum = 2(5 + 5 + 3d)

Sum = 2(10 + 3d)

Now,

Sum = 4/2{(5 + 4d) + (5 + 7d)}

Sum = 2{(5 + 4d) + (5 + 7d)}

Sum = 2{5 + 5 + 4d + 7d}

Sum = 2{10 + 11d}

Now

2(10 + 3d) = 2(10 + 11d)/2

Since it's half

20 + 6d = 10 + 11d

20 - 10 = 11d - 6d

10 = 5d

10/5 = d

\dag{ \textsf{ \textbf{ \pink{ \underline{Difference = 2}}}}}†

Difference = 2

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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