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Between I and 3I, m numbesequence is an A. P. and the ratio of 7h and (m-1)t numvalue of m.rs have been nserted in s |
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Answer» Since m numbers are inserted between 1 and 31 then the total number of terms in the A.P. = n =(m+2).The first term = a = 1The last term = = a+(m+2-1)d= 31or, 1+(m+1)d=31or, (m+1)d=30or, d=30/(m+1) ---------------------(1)Now, according to the question,[a+7d]/[a+(m-1)d]=5/9or, (1+7d)/{1+(m-1)d}=5/9or, 9+63d=5+5(m-1)dor, 63d-5(m-1)d=5-9or,(63-5m+5)d=-4or, (68-5m){30/(m+1)}=-4or, 30(68-5m)=-4(m+1)or, 2040-150m=-4m-4or, -150m+4m=-2040-4or, -146m=-2044or, m=(-2044)/(-146)or, m=14 Ans. hit like if you find it useful |
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