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# each of the series 3+5+7+..... and 4+7+10.......... is continued to 100 terms find how many terms are identical

19 Points
8 years ago

48 terms would be common to both the series...

first take common terms in both series..7, 13,.....so,this has a common difference of 4 starting with 7.

now last term of the given series are 201 and 301 respectively..So we can end our calculation by checking upto 201.

201=7+(n-1)4----->n=48(we have to take integer nearest to it)

vishwa kiran
17 Points
8 years ago

the answer is 33 ..............if the answer is correct then inform me i will explain if u need it and if it correct

Manas Satish Bedmutha
22 Points
8 years ago
There are 33 such no.s, I agree. Method I: The reason being, consider commons in both series - 7,13, ... till 199. For 1st & 2nd series resp. last terms are 199 and 299. Thus to find commons in no.s till 199 such that they are 6n+1, where n belongs to natural no.s, thus last is 199. Now if 199 = 6n+1, n=33. Thus the reqd. no. Method II: Add 1 to both series. Thus first term is same. Let ''x''th term of 1st equal to ''y''th of 2nd, for series of 101 terms, as both have 1 added. Therefore, 1+2x = 1+3y. Implies y=(2/3)x. Since x is a natural no. 3 divides x completely. Tus reqd. no. is all evens less than 1 +200=201 such that they are divisible by 3. So reqd. no. = [201/6], [] denotes greatest integer function. Thus required no = 33.