Question

how many terms are there in AP. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

Answer 1

hey

here is answer

given

a1=-14

and a5=2

and

Sn=40

now

a+4d=2... (1) (since a5=2).


now.

a=-14.....(2)


from equation (1) and (2).

a+4d=2
a+0d=-14
_________

4d=16

d=16/4

d=4

means common difference is 4


now

Sn= n/2(2a+(n-1)d)

40=n/2(2*-14+(n-1)4)

40*2=n(-28+4n-4)

80=n(-32+4n)

4n^2-32n=80

4n^2-32n-80=0 (both side ÷ by 4)

n^2-8n-20=0

n^2-10n+2n-20

n(n-10)+2(n-10)

(n-10)(n+2)


now

n-10=0

n=10

and

n+2=0

n=-2

answer will always be in positive so


answer is

$\bf{10}$




hope it helps

thanks

Answer 2

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


$\textsf{Let the A.P. has n terms and common difference is d.}$


$\underline{\textsf{Given, }}$


$\mathsf{\implies First \: term(t_1) \: = \: -14 } \\ \\ \mathsf{\implies Fifth \: term( t_5) \: = \: 2 } \\ \\ \mathsf{\implies S_n \: = \:40 }$



$\underline{\textsf{Now,}} \\ \\ \mathsf{\implies t_1 \: = \: -14 } \\ \\ \mathsf{\implies a \: = \: -14 \qquad...(1) } \\ \\ \textsf{And,} \\ \\ \mathsf{\implies t_5 \: = \: 2 } \\ \\ \mathsf{\implies a \: + \: ( 5 \: - \: 1 )d \: = \: 2} \\ \\ \mathsf{\implies a \: + \: 4d \: = \: 2 } \\ \\ \textsf{Plug the value of (1) , }$
$\mathsf{\implies -14 \: + \: 4d \: = \: 2 } \\ \\ \mathsf{\implies 4d \: = \: 2 \: + \: 14} \\ \\ \mathsf{\implies 4d \: = \: 16 } \\ \\ \mathsf{\implies d \: = \: 16 \: \div \: 4 } \\ \\ \mathsf{\: \therefore \: d \: = \: 4 }$
$\underline{\textsf{Again,}} \\ \\ \mathsf{\implies S_n \: = \: 40 } \\ \\ \mathsf{\implies \dfrac{n}{2}\{2a \: + \: ( n \: - \: 1 )d \} \: = \: 40} \\ \\ \textsf{Plug the value of a and d , }$
$\mathsf{ \implies \dfrac{n}{2} \{2 \: \times \: (- 14) \: + \: (n \: - \: 1)4 \} \: = \: 40 } \\ \\ \mathsf{ \implies \dfrac{n}{2} \{ - 28 \: + \: 4n \: - \: 4 \} \: = \: 40 } \\ \\ \mathsf{ \implies n( 4n \: - \: 32) \: = \: 40 \: \times \: 2 } \\ \\ \mathsf{ \implies4n( n \: - \: 8) \: = \: 80} \\ \\ \sf \implies n(n \: - \: 8) \: = \: 80 \: \div \: 4 \\ \\ \sf \implies {n}^{2} \: - 8n \: = \: 20 \\ \\ \sf \implies {n}^{2} \: - \: 8n \: - \: 20 \: = \: 0$
$\sf \implies {n}^{2} \: - \: 10n \: + \: 2n \: - \: 20 \: = \: 0 \\ \\ \sf \implies n(n \: - \: 10) \: + \: 2( n \: - \: 10) \: = \: 0 \\ \\ \sf \implies (n \: - \: 10)(n \: + \: 2) \: = \: 0$
$\textsf{By Zero Product Rule ,} \\ \\ \sf \implies n \: - \: 10 \: = \: 0 \quad \implies n \: + \: 2 \: = \: 0 \\ \\ \sf \: \: \: \: \: \: \: \: \therefore \: n \: = \: 10 \: \: \qquad \qquad \therefore \: n \: = \: - 2$
$\textsf{But number of terms can't be negative , } \\ \textsf{so n can't be equal to -2 , hence possible } \\ \textsf{value of n is 10.}$