Question

solve the Equation for
n
tn = 148
1+6+11 etc...​

Answer 1

Answer:

Given 1 + 6+ 11+ 16 + ............+x = 148

take the AP :1,6,11,16,..........,x

In this AP

a = 1

d = 6-1 = 5

Given Sn = 148

we 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= -37/5

As n is the no.of terms in the AP can not be fractional   and negative  

∴n=8

Theirfore x is the 8th term

Tn = T8 = a + (n-1)d  

             = 1 + (8-1)5

             = 1+7(5)

             = 1+35

             = 36

∴8th term = x = 36.

i think this will help u mate