Question

find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively?
Plz give fast answer

Answer 1

a2=14 i.e. a+d=14...a=14-d..
a3=18 i.e. a+2d=18..
i.e.14-d+2d=18....14+d=18
d=4...now a=10
now you can easily find the sum of 51 terms by applying a and d in the formula....n/2(2a+(n-1)d)
hope...it helps...

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let\:a\:be\:first\:term\:and\:D\:be\:Common\:difference$

$\bf\huge a_{2} = 14 \:and\: a_{3} = 18$

$\bf\huge a + d = 14 , a + 2d = 18$

$\bf\huge By\:Solving\:Equation$

$\bf\huge d = 4 \:and\: a = 10$

$\bf\huge Substitute\: a = 10 , d = 4 \:and\: n = 51$

$\bf\huge S_{n} = \frac{N}{2}[2a + (n - 1)d]$

$\bf\huge S_{51} = \frac{51}{2}[2\times 10 + (51 - 1)\times 4]$

$\bf\huge = \frac{51}{2}[20 + 50\times 4]$

$\bf\huge = \frac{51}{2}(20 + 200)$

$\bf\huge = \frac{51}{2}\times 220$

$\bf\huge = 51\times 110$

$\bf\huge = 5610$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$