Question

Find the first term of gp whose common ratio is 3 last term is 486 and sum of whose terms is 728

Answer 1

The sequence of gp is $a, ar, ar^{2}, ar^{3},....... ar^{n-1}$

The common ratio(r)=3 (given)

The last term of sequence = 486

$ar^{n-1}=486$

$a(3)^{n-1}=486$

( By using law of exponent, $a^{m} \times a^{n}= a^{m+n}$)

$a(3^{n} \times 3^{-1})=486$

$\frac{a(3^{n})}{3}=486$

${a(3^{n})}=486 \times 3$

${a(3^{n})}=1458$ (Equation 1)

Since, sum of sequence = 728

Sum of gp when (r >1)= $\frac{a(r^{n}-1)}{r-1}$

$\frac{a(r^{n}-1)}{r-1} = 728$

$\frac{a(3^{n}-1)}{3-1}=728$

${a(3^{n}-1)}= 1456$

${a(3^{n})-a}= 1456$

Substituting value of ${a(3^{n})$ from equation 1.

$1458-a=1456$

Therefore, a=2.

Answer 2

Step-by-step explanation:

The sequence of gp is a, ar, ar^{2}, ar^{3},....... ar^{n-1}a,ar,ar2,ar3,.......arn−1

The common ratio(r)=3 (given)

The last term of sequence = 486

ar^{n-1}=486arn−1=486

a(3)^{n-1}=486a(3)n−1=486

( By using law of exponent, a^{m} \times a^{n}= a^{m+n}am×an=am+n )

a(3^{n} \times 3^{-1})=486a(3n×3−1)=486

\frac{a(3^{n})}{3}=4863a(3n)=486

{a(3^{n})}=486 \times 3a(3n)=486×3

{a(3^{n})}=1458a(3n)=1458 (Equation 1)

Since, sum of sequence = 728

Sum of gp when (r >1)= \frac{a(r^{n}-1)}{r-1}r−1a(rn−1)

\frac{a(r^{n}-1)}{r-1} = 728r−1a(rn−1)=728

\frac{a(3^{n}-1)}{3-1}=7283−1a(3n−1)=728

{a(3^{n}-1)}= 1456a(3n−1)=1456

{a(3^{n})-a}= 1456a(3n)−a=1456

Substituting value of {a(3^{n}) from equation 1.

1458-a=14561458−a=1456

Therefore, a=2.