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Which term of the AP 5, 15, 25,… will be 130 more than its 31st term? |
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Answer» AP is 5, 15, 25,… First term = a = 5 Common difference = d = 15 – 5 = 10 Find 31st term: a31 = a + (n – 1)d = 5 + (31 – 1) 10 = 5 + 30 x 10 = 305 Required term = 305 + 130 = 435 Now, say 435 be the nth term, then an = a + (n – 1)d 435 = 5 + (n – 1)10 435 – 5 = (n – 1)10 n – 1 = 43 n = 44 The required term will be 44th term. |
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