Find the Sum of First 51 terms of an AP Whose Secound and Third terms are 14 and 18 Respectively | Bzziii.com
Find the sum of first 51 terms of an AP whose secound and third terms are 14 and 18 respectively.
We Know that
From (1) & (2)
14 - d = 18 - 2d
2d - d = 18 - 14
d = 4
Putting value of d in (1)
a = 14 - d
a = 14 - 4
a = 10
Now, we need to find sum of first 51 terms
According formula,
`"S"_n` = `n/2` [2a + (n - 1) d]
Now, Putting values, (n = 51, a = 10 & d = 4 )
`"S"_n` = `51/2` [2`\times`10 + (51 - 1) 4]
`"S"_n` = `51/2` [20 + 50`\times`4]
`"S"_n` = `51/2` [20 + 200]
`"S"_n` = `51/2` `\times` 220
`"S"_n` = 51 `\times` 110
`"S"_n` = 5610
Hence, the sum of first 51 terms is 5610
`"a"_n` = a + (n - 1) d
Given 2nd term is 14 `"a"_2` = a + (2 - 1) d 14 = a + d 14 - d = a a = 14 - d .....(1) |
Given 3rd term is 18 `"a"_3` = a + (3 - 1) d 18 = a + d 18 - d = a a = 18 - d .....(2) |
14 - d = 18 - 2d
2d - d = 18 - 14
d = 4
Putting value of d in (1)
a = 14 - d
a = 14 - 4
a = 10
Now, we need to find sum of first 51 terms
According formula,
`"S"_n` = `n/2` [2a + (n - 1) d]
Now, Putting values, (n = 51, a = 10 & d = 4 )
`"S"_n` = `51/2` [2`\times`10 + (51 - 1) 4]
`"S"_n` = `51/2` [20 + 50`\times`4]
`"S"_n` = `51/2` [20 + 200]
`"S"_n` = `51/2` `\times` 220
`"S"_n` = 51 `\times` 110
`"S"_n` = 5610
Hence, the sum of first 51 terms is 5610
Comments
Post a Comment