Question

The sum of a few terms of any ratio series is 728, if the common ratio is 3 and the last term is 486, then what will be the first term of series?

Answer 1

Step-by-step explanation:

nth term of series = arn^−1 = a(3)n^−1 = 486 —– (i)

Sum of n terms of series is Sn= a (3n−1) / [3−1] = 728 (∵r >1)

From (i), a (3n / 3) = 486 or [a * 3n] = 3 × 486 = 1458

From (ii), [a * 3n] − a = 728 * 2 or [a * 3n] − a

1458 − a = 1456

or a = 2

Hope it will help you:-)

Answer 2

Answer:

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Step-by-step explanation:

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