Question

find the sum of 51 term of AP in which second and third term are 14 and 18 respectively

Answer 1

Here's your answer mate,

Given that , t2 = 14 & t3 = 18 .

»tn = a + (n-1) d

»t2 = a + (14-1) d

t2 = a + 13d

14 = a + 13d ..............(1)

»t3 = a + (18-1) d

t3 = a + 17d

18 = a + 17d ..............(2)

Subtract equation (1) from (2),

18 = a + 17d..............(2)

-14 = a + 13d..............(1)

_________________

4 = 4d

.·.d = 4/4

.·.d = 1

Put value of d in equation (1)

14 = a + 13(1)

14 = a + 13

.·.a = 14 -13

.·.a = 1

Now we know the values of a and d ,

Sn = n/2 * [ 2a + (n-1) d ]

S51= 51/2 * [ 2*1+(51-1) * 1 ]

S51= 51/2* [2 + 50 ]

S51= 51/2* 52

S51=2652/2

.·.S51= 1326

Problem solved !!!

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge = 51\times 110$

$\bf\huge = 5610$

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