Question

ow many terms of the ap 1 4 7 are needed to make the sum 51

Answer 1

Answer:

first term (a)=1

common difference (d)=3

we know

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

=>51=n/2 {2.1+(n-1) 3}=n/2 {3n-1}

=>3n^2-n=102

=> 3n^2-n-102=0

use quadratic formula

n=6,-17/3

but n =-17/3 is not possible

so, n=6

Answer 2

Answer :

$\large{\star\:\:\boxed{\sf{S\sf{_6}\:=\:51}}\:\:\star}$

Explanation :

Given :–

• 1 , 4 , 7 , 10 , ... are in A.P.

• where a = 1 and d = 3 .

To Find :–

• How many terms(n) will give a Sum of 51 (Sₙ)  .

Formula Applied :–

• $\boxed{\star\:\:\bf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] }\:\:\star}$

Solution :–

We have ,

• a = 1
• d = 3
• Sₙ = 51

Putting these values in the Formula :

$\rightarrow\sf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] }$

$\rightarrow\sf{51\:=\:\dfrac{n}{2}\:[2(1)\:+\:(n\:-\:1)(3)] }$

$\rightarrow\sf{51\:\times\:2\:=\:n\:(2\:+\:3n\:-\:3) }$

$\rightarrow\sf{102\:=\:n\:(3n\:-\:1) }$

$\rightarrow\sf{102\:=\:3n^2\:-\:n }$

$\rightarrow\sf{3n^2\:-\:n\:-\:102\:=\:0 }$

$\rightarrow\sf{3n^2\:-\:18n\:+\:17n\:-\:102\:=\:0 }$

$\rightarrow\sf{3n(n\:-\:6)\:+\:17(n\:-\:6)\:=\:0 }$

$\rightarrow\sf{(3n\:+\:17)(n\:-\:6)\:=\:0 }$

$\rightarrow\sf{n\:=\:6\:,\:(-\dfrac{17}{3}) }$

Since the Number of terms (n) can't be negative !

So , the Positive value will be correct , i.e.

$\rightarrow\boxed{\bf{n\:=\:6 }}$

∴ 6 Terms of this A.P. will give a Sum of 51 .