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Hint : In order to solve this problem treat this series as an AP whose sum is 148 and common difference 5 and first term as 1.

The given series is

1 + 6 + 11 +….+ x = 148

We will treat it as an AP with first term, common difference, sum, and last term as 1,5,148 and x.

Here we have to find x.

As we know,

a = 1

d = 5

${{\text{S}}_n}$ = 148

Last term = x

We know the formula,

Last term = x = a + (n-1)d

So, by putting the values we have in above equation we get

x = 1 +5n – 5

x = 5n – 4

5n = x + 4

Therefore,

${\text{n = }}\dfrac{{{\text{x + 4}}}}{5}$ ……(i)

We have ${{\text{S}}_n}$ = 148,

And we know,

$

{{\text{S}}_n} = \dfrac{n}{2}({\text{a + last term}}) \\

{{\text{S}}_n} = \dfrac{n}{2}({\text{a + x}}) \\

$

From (i) we can say,

$148 = \dfrac{{{\text{x }} + \,\,4}}{{2(5)}}(1{\text{ + x}})$

Solving further we get,

$

1480 = {\text{(x + 4)(x + 1)}} \\

{\text{1480 = (}}{{\text{x}}^2}{\text{ + 5x + 4)}} \\

{{\text{x}}^2}{\text{ + 5x - 1476 = 0}} \\

$

Calculating the roots of above equation bye the formula,

$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

So, we get,

$

{\text{x = }}\dfrac{{ - 5 \pm \sqrt {{5^2} - 4(1)( - 1480)} }}{{2(1)}} \\

{\text{x = }}\dfrac{{ - 5 \pm 77}}{2} \\

$

On solving it further we get ,

x = 36 when we take +77

x = -41 when we take -77

On observing the series the last term of it can never be negative so the one and only value of x is 36.

Therefore, x = 36.

Note – In these types of problems of series first we have to consider which series is given here, it may be AP, GP, HP etc. Then after considering, obtain the specific value with respect to the series. As here in this question we have considered the series as an AP whose first term, common difference, number of terms and sum is given and asked to find the value of last term which was given as a variable. Then after applying the general formula of nth term and sum of an AP we got a quadratic equation in terms of last term solving that gave us the value of last term.

The given series is

1 + 6 + 11 +….+ x = 148

We will treat it as an AP with first term, common difference, sum, and last term as 1,5,148 and x.

Here we have to find x.

As we know,

a = 1

d = 5

${{\text{S}}_n}$ = 148

Last term = x

We know the formula,

Last term = x = a + (n-1)d

So, by putting the values we have in above equation we get

x = 1 +5n – 5

x = 5n – 4

5n = x + 4

Therefore,

${\text{n = }}\dfrac{{{\text{x + 4}}}}{5}$ ……(i)

We have ${{\text{S}}_n}$ = 148,

And we know,

$

{{\text{S}}_n} = \dfrac{n}{2}({\text{a + last term}}) \\

{{\text{S}}_n} = \dfrac{n}{2}({\text{a + x}}) \\

$

From (i) we can say,

$148 = \dfrac{{{\text{x }} + \,\,4}}{{2(5)}}(1{\text{ + x}})$

Solving further we get,

$

1480 = {\text{(x + 4)(x + 1)}} \\

{\text{1480 = (}}{{\text{x}}^2}{\text{ + 5x + 4)}} \\

{{\text{x}}^2}{\text{ + 5x - 1476 = 0}} \\

$

Calculating the roots of above equation bye the formula,

$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

So, we get,

$

{\text{x = }}\dfrac{{ - 5 \pm \sqrt {{5^2} - 4(1)( - 1480)} }}{{2(1)}} \\

{\text{x = }}\dfrac{{ - 5 \pm 77}}{2} \\

$

On solving it further we get ,

x = 36 when we take +77

x = -41 when we take -77

On observing the series the last term of it can never be negative so the one and only value of x is 36.

Therefore, x = 36.

Note – In these types of problems of series first we have to consider which series is given here, it may be AP, GP, HP etc. Then after considering, obtain the specific value with respect to the series. As here in this question we have considered the series as an AP whose first term, common difference, number of terms and sum is given and asked to find the value of last term which was given as a variable. Then after applying the general formula of nth term and sum of an AP we got a quadratic equation in terms of last term solving that gave us the value of last term.