Question

An arithmetic sequence has a 2nd term equal to 7 and 8th term equal to -23. Find the term of the sequence that has value -183

Answer 1

ANSWER :–

GIVEN :–

• Second term of A.P. = 7

• 8th term of A.P. = -23

TO FIND :–

• The term of the sequence that has value -183 = ?

SOLUTION :–

• We know that nth term of A.P. is –

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

• Here –

$\\ \: \: \: \: { \huge{.}} \: \: { \bold{ a =first \: \: term }} \: \:$

$\\ \: \: \: \: { \huge{.}} \: \: { \bold{ d =common \: \: difference}} \: \:$

$\\ \: \: \: \: { \huge{.}} \: \: { \bold{ n =total \: \: terms}} \: \:$

$\\ \: \: \: \: { \huge{.}} \: \: { \bold{ T_{n}=n \th \: \: term}} \: \:$

• According to the first condition –

$\\ \: \: \implies \: \: { \bold{ T_{2} = 7}} \: \:$

$\\ \: \: \implies \: \: { \bold{a + (2 - 1)d = 7}} \: \:$

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

$\\ \: \: \implies \: \: { \bold{a = 7 - d \: \: \: \: \: \: - - - - eq.(1)}} \: \:$

• According to the second condition –

$\\ \: \: \implies \: \: { \bold{ T_{8} = - 23}} \: \:$

$\\ \: \: \implies \: \: { \bold{a + (8 - 1)d = - 23}} \: \:$

$\\ \: \: \implies \: \: { \bold{a + 7d = - 23}} \: \:$

• Using eq.(1) –

$\\ \: \: \implies \: \: { \bold{7 - d + 7d = - 23}} \: \:$

$\\ \: \: \implies \: \: { \bold{7 + 6d = - 23}} \: \:$

$\\ \: \: \implies \: \: { \bold{ 6d = - 30}} \: \:$

$\\ \: \: \implies \: \: { \bold{ d = - 5}} \: \:$

• Now put the value of 'd' in eq.(1) –

$\\ \: \: \implies \: \: { \bold{a = 7 - ( - 5) }} \: \:$

$\\ \: \: \implies \: \: { \bold{a = 12 }} \: \:$

• Now Let's find the term that has value (-183) –

$\\ \: \implies \: { \bold{ - 183= 12 + (n - 1)( - 5)}} \: \:$

$\\ \: \implies \: { \bold{ - 183 - 12= (n - 1)( - 5)}} \: \:$

$\\ \: \implies \: { \bold{ - 195= (n - 1)( - 5)}} \: \:$

$\\ \: \implies \: { \bold{ (n - 1) = \dfrac{ - 195}{ - 5} }} \: \:$

$\\ \: \implies \: { \bold{ (n - 1) = 39 }} \: \:$

$\\ \: \implies \: { \bold{ n = 40}} \: \:$

Hence , 40th term is -183

Answer 2

Given -

2nd term of AP sequence = 7

8th term of AP sequence= - 23 .

To Find :-

The term which is equal to -183 .

Solution :-

$\underline{\sf{\red{\implies a_n = a_{1} + (n-1)d }}}$

So, using this formula -

$\sf{\implies a_2 = a_1 + d = 7 .......eq \: 1st }$

$\sf{\implies a_8 = a_1 + 7d = - 23....... eq \: 2nd }$

Substracting eq 1st from 2nd

$\sf{\implies a_1 + 7d - a_1 - d = -23 - 7 }$

$\sf{\implies 6d = -30 }$

$\sf{\implies d = - \frac{30}{6} \rightarrow -5}$

Value of common difference is -5 .

Putting this value in eq 1st

$\sf{\implies a_1 - 5 = 7 }$

$\sf{\implies a_1 = 7 + 5}$

$\sf{\implies a_1 = 12 }$

Value of a1 is 12 .

$\sf{\implies -183 = a_1 + (n-1)d }$

$\sf{\implies -183 = 12 + (n-1)-5}$

$\sf{\implies -183 = 12 -5n + 5 }$

$\sf{\implies -183 = 17 - 5n }$

$\sf{\implies -183 - 17 = - 5n }$

$\sf{\implies 200 = 5n }$

$\sf{\implies n = \frac{200}{5} \rightarrow 40 }$

So 40th term of the given AP sequence is - 183.