Question

If a1,a2,a3.... are in AP. Such that a8:a5=3:2, then a17:a23 is​

Answer 1

Step-by-step explanation:

a8 = a + 7d

a5 = a + 4d

<math> a8:a5 = (a+7d)/(a+4d) = 3/2

a+7d = 3 => a = 3-7d

so, a17:a23 = (3-7d+16d)/(3-7d+22d)

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Answer 2

Answer:

The value is (3-7d+16d)/(3-7d+22d)

Step-by-step explanation:

What is meant by Arithmetic  progression?

The difference between any two consecutive integers in an arithmetic progression (AP) sequence of numbers is always the same amount. It also goes by the name Arithmetic Sequence.

If there is always the same difference between any two succeeding integers in a sequence, then sequence is said to be in arithmetic progression (AP). For illustration: 1, 3, 5, 7, 9, and 11 in Series 1. In this series, there is always a 2 point difference between any two consecutive integers.

Given,

a8 = a + 7d

a5 = a + 4d

a8:a5 = (a+7d)/(a+4d) = 3/2

a+7d = 3 => a = 3-7d

so, a17:a23 = (3-7d+16d)/(3-7d+22d)

Therefore the value is (3-7d+16d)/(3-7d+22d)

To learn more about Arithmetic Progression refer to :

https://brainly.in/question/11357684

https://brainly.in/question/50145673

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