Question

solve the question no.15

Answer 1

Hope this answer might help you dear

Answer 2

Solution:

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Given:

AP: 5,15,25....

∴ a = 5,

∴ $d = a_2 - a_1 = 15 - 5 = 10$

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To find :

The term which is 130 more than it's 31st term,..

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As, we know that,

 $a_n = a + (n-1)d$

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So,

=> $a_{31} = 5 + (31-1)(10)$

=> $a_{31} = 5 + (30)(10)$

=> $a_{31} = 5 + 300$

=> $a_{31} = 305,.$

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130 more than 31st term

=> $130 + 305 = 435$

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The number of the term be n,

=> $a_n = a+ (n-1)d$

=> $435 = 5 + (n-1)(10)$

=> $435 - 5 = 10n - 10$

=> $430 = 10n- 10$

=> $430 + 10 = 10n$

=> $440 = 10n$

=> $n = 44$

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           ∴ 44th term of the AP is 130 more than it's 31st term,.

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                                        Hope it Helps!!

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