Question

Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of

the next four terms.​

Answer 1

Given

$\tt \to \: first \: term(a) = 5$

$\tt \to \: Sum \: of \: the \: first \: four \: terms \: is \: half \: the \: sum \: of \: the \: next \: four \: terms$

To Find Common Difference (d)

According to question we can write as

$\to \displaystyle \sum _{ \sf \: n = 1} ^{4} \sf \: a _{n} = \frac{1}{2} \sum_{ \sf \: n = 5} ^{8} \sf \: a _{n}$

We can write as

$\tt \to \: a_1 + a_2 + a_3 + a_4 = \dfrac{1}{2} (a_5 + a_6 + a_7 + a_8)$

Use This Formula

$\tt \to \: T_n = a + (n - 1)d$

We get

$\tt \to \: a + (a + d) + (a + 2d) +( a + 3d) = \dfrac{1}{2} (a + 4d + a + 5d + a + 6d + a + 7d)$

$\tt \to \: 4a + 6d = \dfrac{1}{2} (4a + 22d)$

$\tt \to \: 4a + 6d = \dfrac{1}{2} \times 2(2a + 11d)$

$\tt \to \: 4a + 6d = (2a + 11d)$

Now Put a = 5

$\tt \to \: 4 \times 5 + 6d = (2 \times 5 + 11d)$

$\tt \to \: 20+ 6d = 10 + 11d$

$\tt \to \: 20 - 10 = 11d - 6d$

$\tt \to \: 10 = 5d$

$\tt \to \: d = \dfrac{10}{5} = 2$

Answer

Common Difference (d) = 2

Answer 2

Question:-

• Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of  the next four terms.

To Find:-

• Find the common difference.

Solution:-

Formula to be Used:-

• $\sf \: T_n = a + ( n - 1 )d$

We can also write as:-

$\longrightarrow\sf \: a_1 + a_2 + a_3 + a_4 = \dfrac { 1 } { 2 } ( a_5 + a_6 + a_7 + a_8 )$

$\longrightarrow\sf \: a + ( a + d ) + ( a + 2d ) + ( a + 3d ) = \dfrac { 1 } { 2 } ( a + 4d + a + 5d + a + 6d + a + 7d )$

$\longrightarrow\sf \: 4a + 6d = \dfrac { 1 } { 2 } ( 4a + 22d )$

$\longrightarrow\sf \: 4( 5 ) + 6d = \dfrac { 1 } { 2 } \times 2 \times ( 2( 5 ) + 11d )$

$\longrightarrow\sf \: 20 + 6d = 10 + 11d$

$\longrightarrow\sf \: 20 - 10 = 11d - 6d$

$\longrightarrow\sf \: 5d = 10$

$\longrightarrow\sf \: d = \cancel\dfrac { 10 } { 5 }$

$\longrightarrow\sf \: d = 2$

Hence ,

• Common difference is 2