Question

(x) given l = 28, S= 144, and there are total 9 terms. Find a.


by sn=n/2(2a+(n-1)d)​

Answer 1

ANSWER:

• The value of a = 4.

GIVEN:

• L = 28 and S = 144

• There are total 9 terms.

TO FIND:

• The value of a.

EXPLANATION:

$\red{\boxed{ \bold{ \large{ \blue{S_n=\dfrac{n}{2}(2a+(n-1)d)}}}}} \\ \\ \mapsto\sf L = a + (n - 1)d = 28 \\ \\ \mapsto\sf n = 9 \\ \\ \mapsto\sf S_n = 144 \\ \\\mapsto \sf 144 = \dfrac{9}{2} (a + 28) \\ \\ \mapsto\sf 16 = \dfrac{1}{2} (a + 28) \\ \\ \mapsto\sf 32 = a + 28 \\ \\ \mapsto\sf a = 4$

Hence the value of a = 4.

Answer 2

Question :

• Given l = 28, S= 144, and there are total 9 terms. Find a.

Given :

• l = 28
• S = 144
• n = 9

To find :

• First term(a) ?

Required Answer :

• First term(a) of A.P is 4

Step by step explanation :

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Using formula :

$\qquad\red\bigstar\:{\underline{\boxed{\bf{\green{S_{n} = \dfrac{n}{2}[2a + (n - 1)d]}}}}}$

$:\implies\qquad\tt S_{n} = \dfrac{n}{2}[a + a + (n - 1)d]$

$:\implies\qquad\tt S_{n} = \dfrac{n}{2}(a + a_{n})$

$:\implies\qquad\tt S_{n} = \dfrac{n}{2}(a + l)$

Putting all known values :

$:\implies\qquad\tt 144 = \dfrac{9}{2}(a + 28)$

$:\implies\qquad\tt \dfrac{144}{9} = \dfrac{1}{2}(a + 28)$

$:\implies\qquad\tt \dfrac{\cancel{144}}{\cancel{9}} = \dfrac{a + 28}{2}$

$:\implies\qquad\tt 16 = \dfrac{a + 28}{2}$

$:\implies\qquad\tt 16\:\times\:2 = a + 28$

$:\implies\qquad\tt 32 = a + 28$

$:\implies\qquad\tt a = 32 - 28$

$:\implies\qquad{\boxed{\frak{\green{a = 4}}}}\:\red\bigstar$

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$\therefore\:{\underline{\frak{First\:term(a)\:of\:an\:A.P\:\leadsto\:4}}}$