Question

Solve the eqn: 1 + 6 + 11 + 16 + ... + x = 148 . please fast

Answer 1

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.

Answer 2

Answer:

x = 36

Step-by-step explanation:

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

To Find:- Value of x

Solution:-  This is an A.P. with first term = 1 and common difference = 5.

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

⇒ 148 = $\frac{n}{2}$ [2 + (n-1)5]

⇒ 148 × 2 = n [2 + 5n - 5]

⇒ 296 = 5$n^{2}$ - 3n

⇒ 5$n^{2}$ - 3n - 296 =0

⇒ 5$n^{2}$ - 40n - 37n - 296 = 0

⇒ 5n(n-8) - 37(n-8) = 0

⇒ (n-8)(5n-37) = 0

∴ n = 8  or n = $\frac{37}{5}$

Now,

∵ $t_{n}$ = a + (n-1)d

⇒ x = 1 + (8-1)5

⇒ x = 36

∴ x = 36

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