Question

The sum of 4
th and 8
th
terms of an A. P. is 24 and the sum of 6
th and 10
th
terms is 34. Find the first term
and the common difference of the A. P​

Answer 1

Correct Question:-

The sum of 4th and 8th terms of an A. P. is 24 and the sum of 6th and 10th terms is 34. Find the first term and the common difference of the A. P

Answer:-

$\red{\bigstar}$ First term $\large\leadsto\boxed{\tt\green{a = \dfrac{-1}{2}}}$

$\red{\bigstar}$ Common difference $\large\leadsto\boxed{\tt\green{d = \dfrac{5}{2}}}$

• Given:-

• Sum of 4th and 8th term of an A.P is 24.

• Sum of 6th and 10th term of an A.P is 34.

• To Find:-

• First term of the A.P

• Common difference of the A.P

• Solution:-

• According to the question:-

★ The sum of 4th and 8th term of an A.P is 24.

Hence,

➪ $\bf a_4 + a_8 = 24 \dashrightarrow\bf\red{[eqn.i]}$

Also,

★ Sum of 6th and 10th term of an A.P is 34.

Hence,

➪ $\bf a_6 + a_{10} = 34 \dashrightarrow\bf\red{[eqn.ii]}$

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{a_n = a+(n-1)d}}}$

• Substituting eqn.[i] in above Formula:-

➪ $\sf a_4 + a_8 = 24$

➪ $\sf a + (4-1) d + a + (8-1) d = 24$

➪ $\sf a + 3d + a + 7d = 24$

➪ $\sf 2a + 10d = 24$

➪ $\bf a + 5d = 12\dashrightarrow\bf\red{[eqn.iii]}$

• Substituting eqn[ii] in above Formula:-

➪ $\sf a_6 + a_{10} = 34$

➪ $\sf a + (6-1) d + a + (10-1) d = 34$

➪ $\sf a + 5d + a + 9d = 34$

➪ $\sf 2a + 14d = 34$

➪ $\bf a + 7d = 17\dashrightarrow\bf\red{[eqn.iv]}$

• Subtracting eqn.[iii] from eqn.[iv]:-

➪ $\sf (a+7d) - (a+5d) = 17 - 12$

➪ $\sf a + 7d - a - 5d = 5$

➪ $\sf 2d = 5$

★ $\large{\bf\pink{d = \dfrac{5}{2}}}$

• Substituting value of d in eqn.[iii]:-

➪ $\sf a + 5d = 12$

➪ $\sf a + 5 \times \dfrac{5}{2} = 12$

➪ $\sf a + \dfrac{25}{2} = 12$

➪ $\sf a = 12 - \dfrac{25}{2}$

➪ $\sf a = \dfrac{24 - 25}{2}$

★ $\large{\bf\pink{a = \dfrac{-1}{2}}}$

Therefore, first term and common difference are,

• a = -1/2

• d = 5/2

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\Large{\underline{\boxed{\red{\bf{Question}}}}}}$

The sum of $4^{th}$ and $8^{th}$ terms of an A. P. is 24 and the sum of $6^{th}$ and $10^{th}$ terms is 34. Find the first term

and the common difference of the A.P.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\Large{\underline{\boxed{\red{\bf{Answer}}}}}}$

Given :-

• The sum of $4^{th}$ and $8^{th}$ terms of an A. P is 24.
• The sum of $6^{th}$ and $10^{th}$ terms is 34.

To find :-

• First term of A.P = ?
• Common difference of A.P = ?

Solution :-

● We know that $n^{th}$ term of A.P is given by :-

${\large{\underline{\boxed{ \bf{a _n = a + (n - 1)d}}}}}$

Here

• a = First term
• $a_n$ = $n^{th}$ term of A.P
• d = Common Difference

Also

• A.P is Arithmetic Progression.

Therefore, according to the question:-

● Problem no. 1 :-

[The sum of $4^{th}$ and $8^{th}$ terms of an A. P is 24.]

》$4^{th}$ term can be given by -

• a = a
• n = 4
• d = d

So,

${\bf{\implies{a _n = a + (n - 1)d}}}$

${\bf{\implies{a _4 = a + (4 - 1)d}}}$

${\bf{\implies{a _4 = a + (3)d}}}$

${ \boxed{\bf{\implies{a _4 = a + 3d}}}}$

》$4^{th}$ term can be given by -

• a = a
• n = 8
• d = d

${\bf{\implies{a _n = a + (n - 1)d}}}$

${\bf{\implies{a _8 = a + (8 - 1)d}}}$

${\bf{\implies{a _8 = a + (7)d}}}$

${ \boxed{\bf{\implies{a _8 = a + 7d}}}}$

● Problem no. 2 :-

[The sum of $6^{th}$ and $10^{th}$ terms is 34.]

》$6^{th}$ term can be given by -

• a = a
• n = 6
• d = d

${\bf{\implies{a _n = a + (n - 1)d}}}$

${\bf{\implies{a _6= a + (6- 1)d}}}$

${\bf{\implies{a _6= a + (5)d}}}$

${ \boxed{\bf{\implies{a _6 = a + 5d}}}}$

》$10^{th}$ term can be given by -

• a = a
• n = 10
• d = d

${\bf{\implies{a _n = a + (n - 1)d}}}$

${\bf{\implies{a _{10}= a + (10- 1)d}}}$

${\bf{\implies{a _{10}= a + (9)d}}}$

${ \boxed{\bf{\implies{a _{10 }= a + 9d}}}}$

● Now, it is given that :-

》In problem 1 sum of $4^{th}$ and $8^{th}$ terms of an A. P is 24.

So, adding both the A.P getting from problem 1 :-

${ {\bf{\implies{a_4 + a_8 = 24}}}}$

${ {\bf{\implies{a + 3d + a + 7d = 24}}}}$

${ \boxed{ \blue{\bf{\implies{2a \: + \: 10d = 24}}}}}..............(eq.1)$

》In problem 2 the sum of $6^{th}$ and $10^{th}$ terms is 34.

So, adding the A.P getting from problem 2 :-

${ {\bf{\implies{a_6 + a_{10}= 34}}}}$

${ {\bf{\implies{a + 5d + a + 9d = 34}}}}$

${ \boxed{ \blue{\bf{\implies{2a \: + \: 14d = 34}}}}}..............(eq.2)$

● So, here we have two equations which are

• 2a + 10d = 24 .........(1)
• 2a + 14d = 34 .........(2)

Value of a from equation 1 is

${ \bf{ \implies{2a + 10d = 24 }}}$

${ \bf{ \implies{2a = 24 - 10d }}}$

$\boxed{ \bf{ \implies{a = \frac{ 24 - 10d }{2}}}}$

■ So, Common Difference or d is :-

Putting the value of a in equation will give common difference:-

$\\ {\large{\bf{ \implies{2a + 14d = 34}}}}$

$\\ {\large{\bf{ \implies{2 \bigg( \frac{24 - 10d}{2} \bigg) + 14d = 34}}}}$

$\\ {\large{\bf{ \implies{24 - 10d + 14d = 34}}}}$

$\\ {\large{\bf{ \implies{24 + 4d = 34}}}}$

$\\ {\large{\bf{ \implies{ 4d = 34 - 24}}}}$

$\\ {\large{\bf{ \implies{ 4d = 10}}}}$

$\\ {\large{\bf{ \implies{ d = \frac{10}{4} }}}}$

$\underline {\boxed{\large{\bf{ \red{ \implies{ d = \frac{5}{2} }}}}}}$

Or

$\underline {\boxed {\large{\bf{ \red{ \implies{Common \: \: difference = 2.5 }}}}}}$

■ So, first term of A.P is

First term of A.P is is calculated by putting the value of d in equation 1 :-

$\\ {\large{\bf{ \implies{2a + 10d = 24}}}}$

$\\ {\large{\bf{ \implies{2a + 10 \times \frac{5}{2} = 24}}}}$

$\\ {\large{\bf{ \implies{2a + 5 \times 5 = 24}}}}$

$\\ {\large{\bf{ \implies{2a + 25= 24}}}}$

$\\ {\large{\bf{ \implies{2a = 24 - 25}}}}$

$\\ {\large{\bf{ \implies{2a = - 1}}}}$

$\underline {\boxed {\large{\bf{ \red{ \implies{ a= - \frac{1}{2} }}}}}}$

Or

$\underline {\boxed{\large{\bf{ \red{ \implies{ First \: term = -0.5 }}}}}}$

Final answer :-

• Common difference = 2.5 or 5/2
• First term = -1/2 or - 0.5

———————————————————

More formulas from A.P :-

● an = a + (n − 1) × d

● S = n/2[2a + (n − 1) × d]

● S = n/2 (first term + last term)