Question

A geometric progression has a second term of 12 and a sum to infinity of 54. Find the possible values of the first term of the progression.

Answer 1

Answer:

18 and 36

❃Given❃

• Second term of G.P= 12
• Sum of G.P. to infinity= 54

✏To Find:

• Values of first term of G.P.

♚Solution♚

We know that,

$\sf{General\:term\:of\:G.P.= ar^{n-1}}$

where,

• a is first term of G.P.

• r is common difference of G.P.

• n is no. of terms

So,

$\sf{Second\:term\:of\:given\:G.P.= ar^{2-1}}$

$\sf{Second\:term\:of\:given\:G.P.= ar^{1}=ar}$

$\longrightarrow\sf{ar=12}$

$\longrightarrow\sf{r=\dfrac{12}{a}}----------(1)}$

We also know that,

Sum of G.P. to infinity = $\sf{\dfrac{a}{1-r},where\:r<1}$

$\longrightarrow\sf{\dfrac{a}{1-r}=54}$

From (1),

$\longrightarrow\sf{\dfrac{a}{1-\dfrac{12}{a}}=54}$

$\longrightarrow\sf{a=54(1-\dfrac{12}{a})}$

$\longrightarrow\sf{a=54(\dfrac{a-12}{a})}$

$\longrightarrow\sf{a^{2} =54(a-12)}$

$\longrightarrow\sf{a^{2} =54a-648}$

$\longrightarrow\sf{a^{2} -54a+648=0}$

$\longrightarrow\sf{a^{2} -36a-18a+648=0}$

$\longrightarrow\sf{a(a-36)-18(a-36)=0}$

$\longrightarrow\sf{(a-18)(a-36)=0}$

$\longrightarrow\sf{a=18,36}$

Hence, possible values of first term are 18 and 36.