Question

Find the sum of the first 51 terms of an ap whose second and third terms are 14 and 18 respectively

Answer 1

hope it's right!! I guess....

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge = 51\times 110$

$\bf\huge = 5610$

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