Question

Hey.....
PLEASE ANSWER THIS QUESTION.
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Answer 1

$\huge \mathfrak \orange{answer}$

We have, AP 1,6,11,16,.............x 

a=1 and d = 5 

Let an = x

      a + (n-1)d = x

      1 + 5n-5 = x

     so, 5n - 4 = x

             n = x+4/5 .........................................(1)

We have, Sn = 148

                 n/2[a+an] = 148

                 n/2[1+x] = 148

        so, n[1+x] = 296

             By using (1)

           x+4/5[1+x] = 296

             [x+4](x+1) = 1480

       so, x²+5x+4 = 1480

           so, x²+5x - 1476 = 0 

    so, x² + 41x- 36x - 1476 = 0       

$\blue{by \: splitting \: middle \: term} \ \: :$

    = x(x+41)   -36(x+41)

 so, x = 36,-41

⚠

Answer 2

In given A.P., it can be noted that a = 1, d = 5 and nth term (An) = x.

Given that the sum is 148,

Sn = n/2(a+An) = 148

=> 148 = n/2(1 +x)

=> 296 = n(1+x). -(1)

Also, An = a + (n-1)d

=> x = 1 + (n-1)5

=> (x-1)/5 + 1 = n

=> (x+4)/5 = n -(2).

Substituting value of n from (2) in (1),

=> 296 = (x+4)(1+x)/5

=> 1480 = x(1+x) +4(1+x)

=> x²+x+4x+4-1480=0

=> x² + 5x - 1476 = 0

=> x² + 41x - 36x -1476 =0

=> (x-36)(x+41) = 0

=> x = 36, -41..

But, x can't be negative.... (given A.P. is positive)

=> x= 36