Question

11. Solve: 1 + 6 +11 + 16 +.........+x =148.

Answer 1

Given 1 + 6+ 11+ 16 + ............+x = 148take the AP :1,6,11,16,..........,xIn this APa = 1d = 6-1 = 5Given Sn = 148we know that Sn = n/2[2a+ (n-1)d]⇒n/2 [ 2a +(n-1)d] = 148⇒n[2(1) + (n-1)5 ] = 148 ×2⇒n[ 2 +5n - 5 ] = 296⇒n [ -3 + 5n ] = 296⇒ -3n + 5n² = 296⇒ 5n² - 3n -296 = 0⇒ 5n² - 40n + 37n - 296 = 0 ⇒5n( n - 8) + 37( n - 8) = 0⇒ (n - 8) (5n + 37) = 0 ⇒n-8 = 0                or                 5n +37 = 0⇒n = 8                   or                   n= -37/5As n is the no.of terms in the AP can not be fractional   and negative  ∴n=8Theirfore x is the 8th termTn = T8 = a + (n-1)d               = 1 + (8-1)5              = 1+7(5)              = 1+35              = 36∴8th term = x = 36.

Answer 2

Given series 1 , 6 , 11 , 16 , ...

First term = a = a1 = 1

a2 - a1 = 6 - 1 = 5

a3 - a2 = 11 - 6 = 5

a4 - a3 = 16 - 11 = 5

Therefore ,

a2 - a1 = a3 - a2 = ....= 5 = d

Common difference = d = 5

Given sequence is I. A.P.

Let the sum ' n ' terms = Sn

Sn = 148 ( given )

n/2 [ 2a + ( n - 1 )d ] = 148

=> n[ 2 × 1 + ( n - 1 )5 ] = 296

=> n( 2 + 5n - 5 ) = 296

=> n( 5n - 3 ) - 296 = 0

=> 5n² - 3n - 296 = 0

=> 5n² - 40n + 37n - 296 = 0

=> 5n( n - 8 ) + 37( n - 8 ) = 0

=> ( n - 8 )( 5n + 37 ) = 0

n - 8 = 0 or 5n + 37 = 0

n = 8 or n = -37/5

But n should not be negative.

Therefore n = 8

1 + 6 + 11 + 16 + ....8 terms = 148

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