Question

(ii) 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:

2

Step-by-step explanation:

Sum of first n terms of AP = (n/2) (a + l),    where n is the number, a is first and l is last term,

Here,  a = first term = 5

 For first 4 terms,

Sum = (4/2) (5 + 4th term)

        = 2[5 + (5 + 3d)]

        = 2(10 + 3d)

Similarly, for next 4 terms,

Sum = (4/2) (5th term + 8th term)

         = 2[(5 + 4d) + (5 + 7d)]

         = 2(10 + 11d]

 According to the question: sum of the first four terms is half the sum of the next  four terms.

​⇒ 2(10 + 3d) = 2(10 + 11d)/2

​⇒ 20 + 6d = (10 + 11d)

​⇒ 20 - 10 = 11d - 6d

​⇒ 10 = 5d

​⇒ 2 = d = common differnce

Answer 2

Answer:

Given :-

First Term = 5

Sum of first 4 terms = half of next two terms

To Find :-

Common Difference

Solution :-

We know that

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

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

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