Question

how many terms of series 1+6+11+.......must be taken so that their sum is 970​

Answer 1

✰ Answer :

$\large\boxed{\rm{\star\:\:n\:=\:20\:\:\star}}$

✮ Explanation :

❈ Given :–

• A.P.  :- 1 , 6 , 11 , ...
• where a = 1 , d = a₂ - a₁ = 6 - 1 = 5 , Sₙ = 970 .

❈ To Find :–

Number of terms which will make a Sum of 970 . (n)

❈ Formula Applied :–

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

❈ Solution :–

We have , a = 1 , d = 5 , Sₙ = 970 .

Putting these values in Formula $\sf{ S_n \:=\:\frac{n}{2} \:[2a\:+\:(\:n\:-\:1\:)\:\times\:d]}$.

$\rightarrow\rm{S_n\:=\:\dfrac{n}{2}\:[2(1)\:+\:(\:n\:-\:1\:)\:\times\:(5)] }$

$\rightarrow\rm{970\:=\:\dfrac{n}{2}\:[2\:+\:(\:n\:-\:1\:)\:\times\:(5)] }$

$\rightarrow\rm{(\:970\:\times\:2\:)\:=\:n\:\times\:[2\:+\:(\:n\:-\:1\:)\:\times\:(5)] }$

$\rightarrow\rm{1940\:=\:n\:\times\:[2\:+\:5n\:-\:5] }$

$\rightarrow\rm{1940\:=\:n\:\times\:[5n\:-\:3] }$

$\rightarrow\rm{1940\:=\:5n^2\:-\:3n}$

$\rightarrow\rm{5n^2\:-\:3n\:-\:1940\:=\:0}$

$\rightarrow\rm{5n^2\:-\:100n\:+\:97n\:-\:1940\:=\:0}$

$\rightarrow\rm{5n\:(\:n\:-\:20\:)\:+\:97\:(\:n\:-\:20\:)\:=\:0}$

$\rightarrow\rm{(\:5n\:+\:97\:)\:(\:n\:-\:20\:)\:=\:0}$

$\rm{\rightarrow{\:n\:=\:\dfrac{-97}{5}\:,\:20 }}$

✫ We will ignore the value $n=(-\frac{97}{5} )$ because n cannot be negative.

$\therefore\boxed{\rm{n\:=\:20}}$

✦ So , the Sum of 20 terms gives us 970 .

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✬ Additional Information :–

➢ A.P. ( Arithmetic Progression ) is a Series of terms where there is a common difference after each term which is denoted by d.

➣ I have used middle term splitting for finding n (number of terms) but we can also use other alternatives also like Quadratic Formula(Sridharacharya formula) or Completing Square method.

Answer 2

Here is your answer 

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↑ 

a = 3

d = 4

S_n = 1176

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

= n/2(6+(n-1)4)

= n/2(6+4n-4)

1176 = (2n+4n^2)/2

= n/2(2+4n)

2352 = 2n + 4n^2

n = -49/2, n = 24

By solving qudratic equation,

Reject the negative term,

n = 24