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CBSEi Class 10 Sequences (AP and GP) CORE

CBSEi Class 10 Sequences (AP and GP) CORE

CBSEi Class 10 Sequences (AP and GP) CORE

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Here a = 2<br />

6 ( 2) = 3<br />

18 ( 6) = 3<br />

54 ( 18) = 3<br />

So, it is a <strong>GP</strong> with common ratio(r) = 3.<br />

(v) …<br />

Here a =<br />

÷ = × = …<br />

× =<br />

So, it is not a <strong>GP</strong> as ≠ .<br />

From the numerators, it appears as the numbers are in <strong>GP</strong> but in fact they are not.<br />

Example 16: Find the value of k for which 4k + 4, 6k 2, 9k 13 form a <strong>GP</strong>.<br />

Solution: Since 4k + 4, 6k 2 <strong>and</strong> 9k 13 form a <strong>GP</strong>,<br />

therefore, = (=r)<br />

(6k 2) (6k 2) = (4k + 4) (9k 13)<br />

36k 2 24k + 4 = 36k 2 52k + 36k 52<br />

= 36k 2 16k 52<br />

8k = 56<br />

k = 7<br />

55

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