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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 7
Total: 106
an = 4n−3,a17,a24
Solution:  
We have, an = 4n−3,a17,a24
Put n = 17 ⇒ a17 = 4(17) – 3 = 68 – 3 = 65
put n = 24 ⇒ a24 = 4 (24) – 3 = 96 – 3 = 93.
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