Question

The first and last terms of an AP are 4 and 81 respectively. The common difference is 7.

(a) How many terms are there in the AP
(b) Find the sum?


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Answer 1

$\huge\bold{Given :)}$

• First term (a) = 4
• Last tern = 81
• Common Difference (d) = 7

$\:\:\:\:\:\bold{\underline{\sf{\green{By\: Using\: Formula}}}}$

$\dag\:\:\small\boxed{\sf{\blue{a_{n}\:=\:a +(n\:-\:1)d}}}$

$\sf\underline\red{Putting~values ~:)}$

$:\implies\sf\: 81 = 4 + (n - 1)7$

$:\implies\sf\: 81 = 4 + 7n - 7$

$:\implies\sf\: 81 = -3 + 7n$

$:\implies\sf\: 81 + 3 = 7n$

$:\implies\sf\: 84 = 7n$

$:\implies\sf\:n = \cancel\dfrac{84}{7}$

$:\implies\large{\underline{\boxed{\sf{\pink{n\:=\:12}}}}}$

$\bold{\underline{\sf{There\:are\:12\:Terms\:in\;AP.}}}$

$\rule{200}2$

$\sf\underline\red{Now~finding~sum~:)}$

$\:\:\:\:\:\bold{\underline{\sf{\green{Formula}}}}$

$\dag\:\;\small\boxed{\sf{\blue{S_{n}\:=\: \frac{n}{2}[2a \:+\:(n\:-\:1)d]}}}$

$\sf\underline\red{Putting~values ~:)}$

$:\implies\sf\: \cancel\dfrac{12}{2} [2(4) +(12 - 1)7]$

$:\implies\sf\:6[ 8 + 77]$

$:\implies\sf\:6 \times 85$

$:\implies\large{\underline{\boxed{\sf{\red{S_{n}\:=\: 510}}}}}$

$\bold{\underline{\sf{Hence,\:Sum\:of\: Given\:AP\:is \:510.}}}$

Answer 2

Answer:

an=a+(n-1)d

81=4+(n-1)7

81\4=7n-7

21=7n-7

7n=21+7

7n=28

n=4