Question

In an A.P.the first term is -5 and last term is 45.If sum of all numbers in the A.P. is 120,then how many terms are there? What is the common difference?

Answer 1

• First term (a) = - 5

• Last term $(a_{n})$ = 45

• Sum of terms $(S_{n})$ = 120

________________ [GIVEN]

• We have to find number of terms (n) and common difference (d)

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$\boxed{*\:a_{n}\:=\:a\:+\:(n\:-\:1)d}$

Put the known values in above equation

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

$\implies$ 50 = (n - 1)d _______ (eq 1)

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$\boxed{*\:S_{n}\:=\:\dfrac{n}{2}\:(a\:+\:a_{n})}$

Put the known values in above equation

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

$\implies$ 120 × 2 = n(40)

$\implies$ 240 = 40n

$\implies$ n = 6

» Put value of d in (eq 1)

$\implies$ 50 = (6 - 1)d

$\implies$ 50 = 5d

$\implies$ d = 10

_____________________________

• $\huge{n\:=\:6}$

$\bold{Number \:of\:terms}$

• $\huge{d\:=\:10}$

$\bold{Common\: difference}$

______________ [ANSWER]