Question

if the sum of the sequence 1,6,11,16...x is148 then find the value of x

Answer 1

The given series is in an AP
In the above AP:

a=1, d=6-1=5, an=x, Sn=148

Now,Sn=148
Sn=n/2(2a+(n-1)d)
296 =n(2+5n-5')
296=2n+5n^2-5n
5n^2-3n-296=0
5n-40n+37n-296=0
5n(n-40)+37(n-40)
(n-40)(n+37)

Hence, n=40,ignoring n in negative

Now an=x
a40= a +39D
a40=1+39*5
x=196=ANS

PLS MARK AS BRAINLIEST

Answer 2

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

Therefore, 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.

Hope it helps.

please mark this answer as brainliest.