Question

Find a and d for an arithmetic sequence where the 4th term is 9 and the 5th term is 7

Answer 1

GIVEN :–

• 4th term of A.P. = 9

• 5th term of A.P. = 7

TO FIND :–

• Value of First term (a) = ?

• Value of Common difference (d) = ?

SOLUTION :–

• We know that If First term (a) , Common difference (d) and total number of term (n) then nth term –

$\\ \: \: \longrightarrow \: \: \large { \boxed{ \bold{T_{n} = a + (n - 1)d}}}$

• According to the first condition –

$\\ \implies { \bold{T_{4} = 9}}$

$\\ \implies { \bold{a + (4 - 1)d = 9}}$

$\\ \implies { \bold{a +3d = 9}}$

$\\ \implies { \bold{a = 9 - 3d \: \: \: - - - eq.(1)}}$

• According to the second condition –

$\\ \implies { \bold{T_{5} = 7}}$

$\\ \implies { \bold{a + (5- 1)d = 7}}$

$\\ \implies { \bold{a +4d = 7}}$

• Using eq.(1) –

$\\ \implies { \bold{(9 - 3d)+4d = 7}}$

$\\ \implies { \bold{9 + d = 7}}$

$\\ \implies { \bold{ d = 7 - 9}}$

$\\ \implies \large { \boxed{ \bold{ d = - 2}}}$

• Put the value of 'd' in eq.(1) –

$\\ \implies { \bold{a = 9 - 3( - 2) \: }}$

$\\ \implies { \bold{a = 9 + 6 \: }}$

$\\ \implies \large { \boxed{ \bold{ a= 15}}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• 4th term AP = 9
• 5th term of AP = 7

${\bf{\blue{\underline{To\:Find:}}}}$

• a = ?
• d = ?

${\bf{\blue{\underline{Formula\:Used:}}}}$

${\bigstar \: \: {\boxed{\boxed{\green{ \sf{ \: a_{n} = a + (n - 1)d}}}}}}$

• Where an is the nth term of AP.
• a is the first term.
• d is common difference.
• n is number of terms.

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ a_{4} = a + (4 - 1)d}}$

$: \implies{\sf{ a + (4 - 1)d = 9}}$

$: \implies{\sf{ a + 3d = 9......(1)}}$

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$: \implies{\sf{ a_{5} = a + (5- 1)d}}$

$: \implies{\sf{ a_{5} = a + (4)d}}$

$: \implies{\sf{ a + 4d = 7.......(2)}}$

___________________________________

By Substitution method,

From (1),

$: \implies{\sf{ a = 9-3d.....(3)}}$

${\bf{\blue{\underline{Put\:the\:Value\: of\:a\: in\:eq\:(2):}}}}$

$: \implies{\sf{ 9- 3d + 4d = 7}}$

$: \implies{\sf{ 9 + d = 7}}$

$: \implies{\sf{ d = 7 - 9}}$

$: \implies \boxed{\sf{ d = -2}}$

Now put the value of d in (3)

$: \implies{\sf{ a = 9 - 3( - 2)}}$

$: \implies{\sf{ a = 9 + 6}}$

$: \implies\boxed{\sf{ a = 15}}$

Hence, a = 15 and d = -2