Question

The sum of the 4rth and 8th term of an AP is 4 and the sum of 6th and 10th term of AP in 12 . Find the first three terms.

Answer 1

4 th term = a + 3d

8 th term = a + 7d

4th + 8th = 2a + 10 d = 4 --(1)

Similarly

6th + 10th =2a + 14d = 12 ---(2)

(2)-(1)

4d = 8

d = 2

Put d = 2 in (1)

2a + 10(2) =4

2a =4-20

a = -8

Therefore the first three terms are

= a, a+d, a+2d

= -8, -6, -4 =Ans..

Answer 2

$\underline{\mathfrak{ Solution : }}$

Let the first term is $\textsf{\textbf{a}}$ and common difference is $\textsf{\textbf{d}}$.

$\textsf{ Formula to be used : }$

$\boxed{\mathsf{\implies T_n \: = \: a \: + \:( n \: - \: 1)d}}$

$\textsf{ According to question : } \\ \\ \mathsf{\implies T_{4} \: + \: T_{8} \: = \: 4 } \\ \\ \mathsf{ \implies a \: + \: (4 \: - \: 1)d \: + \: a \: + \: (8 \: - \: 1)d \: = \: 4} \\ \\ \mathsf{ \implies 2a \: + \: 3d \: + \: 7d \: = \: 4} \\ \\ \mathsf{ \implies2a \: + \: 10d \: = \: 4} \\ \\ \mathsf{ \implies {2}(a \: + \: 5d) \: = \: {4} \: } \\ \\ \mathsf{\implies ( a \: + \: 5d ) \: = \dfrac{4}{2} } \\ \\ \mathsf{\implies ( a \: + \: 5d ) \: = \:2 } \\ \\ \mathsf{ \therefore \: \: a \: = \: 2 \: - \: 5d \qquad...(1)}$

$\textsf{ Now,} \\ \\ \mathsf{ \implies T_{6} \: + \: T_{10} \: = \:12 } \\ \\ \mathsf{\implies a \: + \: ( 6 \: - \: 1 )d \: + \: a \: + ( 10 \: - \: 1 )d \: = \: 12 } \\ \\ \mathsf{\implies 2a \: + \: 5d \: + \:9d \: = \: 12 } \\ \\ \mathsf{\implies 2a \: + \: 14d \: = \: 12 } \\ \\ \mathsf{\implies {2}( a \: + \: 7d ) \: = \: {12} \: \: } \\ \\ \mathsf{\implies ( a \: + \: 7d ) \: = \: \dfrac{12}{2} } \\ \\ \mathsf{ \implies ( a \: + \: 7d ) \: = \:6 }$

$\textsf{ Substitute the value of ( 1 ), } \\ \\ \mathsf{ \implies 2 \: - \: 5d \: + \: 7d \: = \: 6 } \\ \\ \mathsf{ \implies 2 \: + 2d \: = \: 6 } \\ \\ \mathsf{\implies 2d \: = \: 6 \: - \: 2 } \\ \\ \mathsf{\implies 2d \: = \: 4 } \\ \\ \mathsf{\implies d \: = \dfrac{4}{2} } \\ \\ \mathsf{\therefore \: \: d \: = \: 2 }$

$\textsf{ Substitute the value of \textbf{d} in (1) ,} \\ \\ \mathsf{\implies a \: = \: 2 \: - \: 5d } \\ \\ \mathsf{\implies a \: = \: 2 \: - \: 5 \: \times \: 2 } \\ \\ \mathsf{\implies a \: = \: 2 \: - \: 10 } \\ \\ \mathsf{\therefore \: \: a \: = \: -8 }$

$\textsf{ Now , } \\ \\ \textsf{ First 3 terms : } \\ \\ \mathsf{ = ( a ) \: , \: ( a \: + \: d ) \: , \: ( a \: + \: 2d )} \\ \\ \mathsf{ = ( - 8 )\: \:, \:( -8 \: + \:2 ) \:, \: ( -8 \: + \: 2 \: \times \:2 ) } \\ \\ \mathsf{= ( -8 ) \: , \: ( -6 \: + \: 4 ) \: , \:( -8 \: + \: 4) } \\ \\ \mathsf{=( -8) \: , \:( -6 )\: , \: (-4) }$