Question

Find the sum of 1st 22 terms of an ap in which d is 7 and 22 term is 149

Answer 1

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

$\bigstar\sf\underline\orange{Given:-}$

$\longrightarrow\sf{d = 7}$

$\longrightarrow\sf{ 22 th\:\: term\: = \:149}$

_________________________

⠀

$\longmapsto\sf\red{To\: find\: the \:value \:of\: a }$

$\longrightarrow\sf{an = a + (n - 1)d}$

$\longrightarrow\sf{149 = a + (22 - 1)7}$

$\longrightarrow\sf{149 = a + 21 \times 7}$

$\longrightarrow\sf{149 = a + 147}$

$\longrightarrow\sf{a = 149 - 147}$

$\longrightarrow\sf{a = 2}$

⠀

$\longmapsto\sf\red{To\: find \:the\: sum\: of \:first\: 22 \:terms }$

$\longrightarrow\sf{Sn = \dfrac{n}{2} (a + l) }$

$\longrightarrow\sf{Sn= \dfrac{22}{2} (2 + 149) }$

$\longrightarrow\sf{Sn = 11 \times 151 }$

$\longrightarrow\sf{Sn = 1661}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

Answer 2

$\huge\tt\red{Given→}$

• d(common difference)= 7
• 7a+21d(22nd term)= 149

$\huge\tt\red{To\:\:find→}$

• $s_{22} = ?$

As,

→$a + 21d = 149$

→$a + (21 \times 7) = 149$

→$a = 149 - 147$

→$a = 2$

Now,

→$s_{n} = \frac{n}{2} (2 + l )$

→$s_{n} = \frac{22}{2} (2 + 149)$

→$s_{n} = 11 \times 151$

→$s_{n} = 1661$

$Hence, \: s_{22} = 1661$