Question

the first term of an ap is -2 and the last term is 45 if the sum of terms of a is 120 find the number of term and common difference​

Answer 1

Step-by-step explanation:

Correct Question:-

The first term of an AP is -5 and the last term is 45 if the sum of n terms is 120. Find the number of terms and common difference

$\bf\underline{\underline{\red{Given:-}}}$

• The first term = -5

• Last term = 45

• Sum of n terms = 120

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The number of terms

• Common difference

$\bf\underline{\underline{\green{Solution:-}}}$

As we know that:-

The formula used for finding the sum of n terms of an AP is

$\boxed{ \rm \leadsto S_{n} = \frac{n}{2} \big \{a + l \big \}}$

Here:-

•$\rm S_{n} = sum \: \: of \: n \: terms= 120$

• $\rm n =number\: of \: terms= ?$

• $\rm a=first \: term= -2$

• $\rm l=last \: term = 45$

Now:-

${ \rm \implies S_{n} = \dfrac{n}{2} \big \{a + l \big \}}$

${ \rm \implies 120 = \dfrac{n}{2} \big \{ (- 5)+ (45)\big \}}$

${ \rm \implies 120 = \dfrac{n}{2} \big \{ - 5+ 45\big \}}$

${ \rm \implies 120 = \dfrac{n}{2} \big \{ 40\big \}}$

${ \rm \implies 120 = 20n }$

${ \rm \implies \dfrac{120}{20} = n }$

${ \rm \implies n = 6 }$

$\underline{ \boxed{ \pink{\rm \therefore Number \: of \: terms = 6 }}}$

Now let us find the common difference

The formula used for finding nth term of an AP is:-

${ \boxed{ {\rm \leadsto a_{n} = a + (n - 1)d }}}$

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

${{ {\rm \implies 45= - 5+ 5d }}}$

${{ {\rm \implies 45 + 5= 5d }}}$

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

${{ {\rm \implies d = 10}}}$

$\underline{ \boxed{ \purple{\rm \therefore Common \: difference= 10}}}$