Question

3. The first term of an AP is -5 and the last term is 45. If the sum of the terms of an AP is 120, then find the number of terms and the common difference.

Answer 1

no.of terms = 6

common difference = 10

Answer 2

Answer:

Step-by-step explanation:

❈ Given:-

• a=(-5)    [ First Term ]
• l=45      [ Last Term ]
• Sₙ=120  [ Sum of n Terms ]

❈ To Find:-

• n   [ Number of Terms whose Sum is 120 ]
• d   [ Common Difference ]

❈ Formula Applied:-

• Sₙ=$\frac{n}{2}$[a+l] , where n=number of terms, a=first term, l=Last Term and Sₙ=Sum of n terms.

• aₙ=a+(n-1)d, where a=First Term, n=number of terms, d=common difference and aₙ=Last Term.

❈ Solution:-

First we will find n (number of terms) :

      $S_n=\frac{n}{2}[a+l]\:where,\:S_n=120,\:a=(-5)\:and\:l=45.$

$\implies 120=\frac{n}{2}[-5+45]$

$\implies 120\times 2=n(40)$

$\implies 240=40n$

$\implies n=\frac{240}{40}$

$\implies \star\:\:\boxed{n=6}\:\:\star$

Now we have to find d ( Common Difference ) :

       $a_n=a+(n-1)d,\:where\:a=(-5),\:a_n(l)=45\:and\:n=6.$

$\implies 45=(-5)+(6-1)d$

$\implies 45+5=(5)d$

$\implies 50=5d$

$\implies d=\frac{50}{5}$

$\implies \star\:\:\boxed{d=10}\:\:\star$

∴ The Value of n=6 and d=10.

✰ Extra Information :-

→ There is one more formula  $S_n=\frac{n}{2} [2a+(n-1)d]$  which is same as the the formula  $S_n=\frac{n}{2} (a+l)$.