The first term of an A.P.

Question

The first term of an A.P. of 30 non-negative terms is $\frac{10}{3}$. If the sum of this A.P. is the cube of its last term, then its common difference is:

Accepted Answer

To solve this problem, we will use the standard formulas for an Arithmetic Progression (A.P.) as taught in the NCERT syllabus. Let us break down the information given in the question step by step.

Given:

The number of terms in the A.P. is $n=30$.

The first term is $a=\frac{10}{3}$.

All the terms in the A.P. are non-negative.

The sum of the A.P. is equal to the cube of its last term.

Let the common difference of the A.P. be $d$ and the last term (which is the 30th term) be $l$.

The formula for the $n$-th term of an A.P. is:

$l=a+(n-1)d$

$l=\frac{10}{3}+(30-1)d$

$l=\frac{10}{3}+29d\,$

The formula for the sum of $n$ terms of an A.P. when the first and last terms are known is:

${S}_{n}=\frac{n}{2}(a+l)$

For $n=30$, we substitute our known values:

${S}_{30}=\frac{30}{2}(\frac{10}{3}+l)$

${S}_{30}=15(\frac{10}{3}+l)$

${S}_{30}=50+15l$

According to the problem, the sum of this A.P. is equal to the cube of its last term. So, we can write:

${S}_{30}={l}^{3}$

Now, substitute the value of ${S}_{30}$ into this equation:

$50+15l={l}^{3}$

Rearranging the terms to form a standard cubic equation, we get:

${l}^{3}-15l-50=0$

To find the roots of this cubic equation, we can use the trial and error method for integer values.

Let us check $l=5$:

$(5{)}^{3}-15(5)-50=125-75-50=0$

Since $l=5$ satisfies the equation, $(l-5)$ is a factor.

Now, we can factorise the cubic polynomial completely:

${l}^{3}-5{l}^{2}+5{l}^{2}-25l+10l-50=0$

${l}^{2}(l-5)+5l(l-5)+10(l-5)=0$

$(l-5)({l}^{2}+5l+10)=0$

This gives us two parts:

$l-5=0\, \implies \, l=5$

${l}^{2}+5l+10=0$

For the quadratic equation ${l}^{2}+5l+10=0$, the discriminant is:

$D=(5{)}^{2}-4(1)(10)=25-40=-15$

Since the discriminant is negative ($D$l=5$

Now, substitute $l=5$ back into Equation 1 to find the common difference $d$:

$5=\frac{10}{3}+29d$

Subtract $\frac{10}{3}$ from both sides:

$29d=5-\frac{10}{3}$

$29d=\frac{15-10}{3}$

$29d=\frac{5}{3}$

Now, divide both sides by 29:

$d=\frac{5}{3	imes 29}$

$d=\frac{5}{87}$

Thus, the common difference of the given A.P. is $

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