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Question

Mathematics Question on Arithmetic Progression

The number of terms common between the sum 1+2+4+8+.....1 + 2 + 4 + 8 + .....to 100100 terms and 1+4+7+10+.....1 + 4 + 7 + 10 + .....to 100100 terms is

A

6

B

4

C

5

D

none of these.

Answer

5

Explanation

Solution

For a G.P.1+2+4+8+.....Tn=2n1G.P. 1+2+4+8+.....T_{n} = 2^{n-1} For an A.P.1+4+7+10+.....A.P. 1+4+7+10+..... Tm=1+(m1)3=3m2T_{m} = 1+\left(m-1\right)3 =3m-2 They are common if 2n1=3m2or2n1+2=3m2^{n-1} = 3m-2\, or \,2^{n-1} +2 = 3m i.e., 2n2+1=3m232(100)=1502^{n-2} +1 = \frac{3m}{2}\le\frac{3}{2}\left(100\right) = 150 n9,m100\Rightarrow n\le9, m\le100 By trial, n=1,m=1;n=3,m=2;n=5,n=1, m=1 ; n=3, m=2 ; n=5, m=6,n=7,m=22;n=9,m=86 m=6, n=7, m=22; n=9,m=86