Find sum of first 10 term of AP whose 3rd term is 8 and 11th term is 20 .

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Solution

$T_{3} = a + 2 d = 8$ ………. $(1)$
$T_{11} = a + 10 d = 20$ ………….. $(2)$
Equation $(2)$ - Equation $(1)$
$\Rightarrow 8 d = 12$
$d = \frac{12}{8} = \frac{3}{2}$
$a = 8 - 2 d = 8 - 2 (\frac{3}{2})$
$a = 5$
$S_{10} = \frac{10}{2} [2 a + 9 d]$
$= 5 [10 + \frac{27}{2}]$
$= \frac{235}{2}$ .

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Find sum of first 10 term of AP whose 3rd term is 8 and 11th term is 20 .

T3=a+2d=8 ……….(1)T11=a+10d=20 …………..(2)Equation (2) - Equation (1)⇒8d=12d=128=32a=8−2d=8−2(32)a=5S10=102[2a+9d]=5[10+272]=2352.

$T_{3} = a + 2 d = 8$ ………. $(1)$
$T_{11} = a + 10 d = 20$ ………….. $(2)$
Equation $(2)$ - Equation $(1)$
$\Rightarrow 8 d = 12$
$d = \frac{12}{8} = \frac{3}{2}$
$a = 8 - 2 d = 8 - 2 (\frac{3}{2})$
$a = 5$
$S_{10} = \frac{10}{2} [2 a + 9 d]$
$= 5 [10 + \frac{27}{2}]$
$= \frac{235}{2}$ .