Question

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

Answer 1

I think it may help you.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\rightarrow{a_{2}=14}$

a + d = 14 ..... (1)

$\rightarrow{a_{3}=18}$

a + 2d = 18 ...... (2)

★Subtracting Equation (1) from (2)

d = 4

★Substitute the value of d in (1)

a + (4) = 14

a = 14 - 4

a = 10

★Now sum of 51th term :-

${\boxed{\sf\:{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}$

${\boxed{\sf\:{S_{51}=\dfrac{51}{2}[2(10)+(51-1)(4)]}}}$

${\boxed{\sf\:{S_{51}=\dfrac{51}{2}[20+50\times 4]}}}$

${\boxed{\sf\:{S_{51}=\dfrac{51}{2}[20+200]}}}$

${\boxed{\sf\:{S_{51}=\dfrac{51}{2}\times 220}}}$

= 51 × 110

= 5610

${\boxed{\sf\:{Hence\;sum\;of\;n^{th}\;term\;is\;5610}}}$