Question

The sum of a gp with common ratio 3 and the last term is 486 . If the sum of these terms be 728 find the first term

Answer 1

Hey !

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$$a(3)^{n-1}=486$$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 = 728Sum 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.

By FireMaster , moderator , brainly-in team

Answer 2

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 = 486ar^{n-1}=486arn−1=486 a(3)^{n-1}=486a(3)n−1=486 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 = 728Sum 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=1456 Therefore, a=2.

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