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Formula for Sum of n Terms of an AP

In an AP, giv...

Question

In an AP, given $a_{12} = 37, d = 3,$ find 'a' and $S_{12} .$

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Solution

$a + (n - 1) d = a_{n}$

$a + (12 - 1) 3 = 37$

$a + 11 \times 3 = 37$

$a + 33 = 37$

$∴ a = 37 - 33$

$∴ a = 4$

$S_{n} = \frac{n}{2} [a + a_{n}]$

$S_{12} = \frac{12}{2} [4 + 37]$

$= 6 [4 + 37]$

$∴ S_{2} = 246$

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Similar questions

Q. Question 3 (iii)
In an AP:
(iii) Given

a_{12} = 37, d = 3

S_{12}

In an AP

(i) Given $a = 5, d = 3, a_{n} = 50$ , find $n$ and $S_{n}$ .

(ii) Given $a = 7, a_{13} = 35$ , find $d$ and $S_{13}$ .

(iii) Given $a_{12} = 37, d = 3$ , find $a$ and $S_{12}$ .

(iv) Given $a_{3} = 15, S_{10} = 125$ , find $d$ and $a_{10}$ .

(v) Given $d = 5, S_{9} = 75$ , find $a$ and $a_{9}$ .

(vi) Given $a = 2, d = 8, S_{n} = 90$ , find $n$ and $a_{n}$ .

(vii) Given $a = 8, a_{n} = 62, S_{n} = 210$ , find $n$ and $d$ .

(viii) Given $a_{n} = 4, d = 2, S_{n} = - 14$ , find $n$ and $a$ .

(ix) Given $a = 3, n = 8, S_{8} = 192$ , find $d$ .

(x) Given $l = 28, S_{n} = 144$ and there are total $9$ terms. Find $a$ .

Q. In an AP:
(i) Given

a = 5, d = 3

a_{n} = 50

n

S_{n}

a = 7

a_{13} = 35

d

S_{13}

a_{12} = 37

d = 3

a

S_{12}

a_{3} = 15

S_{10} = 125

d

a_{10}

d = 5

S_{9} = 75

a

a_{9}

a = 2, d = 8

S_{n} = 90

n

a_{n}

a = 8

a_{n} = 62

S_{n} = 210

n

d

a_{n} = 4

d = 2

S_{n} = - 14

n

a

a = 3, n = 8, S = 192

d

l = 28, S = 144

, and there are total 9 terms. Find

a

Q. In an AP, given

a_{12} = 37

= 3

S_{12}

a_{12} = 37, d = 3

S_{12}

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