Consider an arithmetic series and a geometric series having four initial terms from the set 11, 8, 21, 16, 26, 32, 4. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q40 with Solution

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Correct answer: 3

Identify the 4-term arithmetic progression (A.P.) and geometric progression (G.P.) from the given set

Given set: $\{11,8,21,16,26,32,4\}$

We need to find four numbers from this set that can be the initial four terms of:

an arithmetic sequence
a geometric sequence

Find the A.P.

Look for 4 numbers with a common difference.

Observe: $11,16,21,26$

This is an A.P. with common difference $d=5$

So the arithmetic series starts as: $11,16,21,26,\dots$

Find the G.P.

Look for 4 numbers with a common ratio.

Observe: $4,8,16,32$

This is a G.P. with common ratio $r=2$

So the geometric series starts as: $4,8,16,32,\dots$

Find the maximum possible four-digit last term in each sequence

We extend each sequence until the last term is the greatest possible 4-digit number appearing in that sequence.

A.P.

General term: $a_n=11+(n-1)5=5n+6$

We want the largest 4-digit term, i.e. $a_n\le 9999$

So, $5n+6\le 9999$ $5n\le 9993$ $n\le 1998.6$

Hence maximum integer $n$ is $n=1998$

Last term: $a_{1998}=11+1997\cdot 5=9996$

So the A.P. is considered up to $9996$

G.P.

General term: $g_n=4\cdot 2^{n-1}$

We want the largest 4-digit term: $4\cdot 2^{n-1}\le 9999$ $2^{n+1}\le 9999$

Now,

\qquad 2^{14}=16384>9999$$ Thus the largest 4-digit term in the G.P. is $$8192$$ This corresponds to $$4\cdot 2^{11}=8192$$ So the G.P. is considered up to $$8192$$ --- 5. **Find common terms of the two series** We need numbers common to both: - A.P.: $$11,16,21,26,31,36,\dots,9996$$ - G.P.: $$4,8,16,32,64,128,256,512,1024,2048,4096,8192$$ A common term must satisfy $$11+5k = 4\cdot 2^m$$ Now check the G.P. terms modulo $5$. Since A.P. terms are $$11,16,21,26,\dots$$ all are congruent to $$1 \pmod 5$$ So common terms must be G.P. terms congruent to $1 \pmod 5$. Now powers of $2$ modulo $5$ cycle as: $$2,4,3,1,2,4,3,1,\dots$$ Thus $$4\cdot 2^{n-1} \pmod 5$$ will cycle through: $$4\cdot 1=4, \ 4\cdot 2=3, \ 4\cdot 4=1, \ 4\cdot 3=2, \dots$$ So we can directly inspect the G.P. terms: $$4,8,16,32,64,128,256,512,1024,2048,4096,8192$$ Among these, - $$16 \equiv 1 \pmod 5$$ - $$256 \equiv 1 \pmod 5$$ - $$4096 \equiv 1 \pmod 5$$ Now verify they are in the A.P.: $$16=11+5(1)$$ $$256=11+5(49)$$ $$4096=11+5(817)$$ All are indeed terms of the A.P. The next such G.P. term would be $$65536$$ which is not 4-digit, so not allowed. Hence the common terms are: $$16,256,4096$$ Therefore, the number of common terms is $$\boxed{3}$$ --- 6. **Compare with stored correct answer** Stored correct answer: $3$ Our derived answer is also $3$, so it agrees.

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View written solutionFree

Correct answer: 3

Identify the 4-term arithmetic progression (A.P.) and geometric progression (G.P.) from the given set

Given set: $\{11,8,21,16,26,32,4\}$

We need to find four numbers from this set that can be the initial four terms of:

an arithmetic sequence
a geometric sequence

Find the A.P.

Look for 4 numbers with a common difference.

Observe: $11,16,21,26$

This is an A.P. with common difference $d=5$

So the arithmetic series starts as: $11,16,21,26,\dots$

Find the G.P.

Look for 4 numbers with a common ratio.

Observe: $4,8,16,32$

This is a G.P. with common ratio $r=2$

So the geometric series starts as: $4,8,16,32,\dots$

Find the maximum possible four-digit last term in each sequence

We extend each sequence until the last term is the greatest possible 4-digit number appearing in that sequence.

A.P.

General term: $a_n=11+(n-1)5=5n+6$

We want the largest 4-digit term, i.e. $a_n\le 9999$

So, $5n+6\le 9999$ $5n\le 9993$ $n\le 1998.6$

Hence maximum integer $n$ is $n=1998$

Last term: $a_{1998}=11+1997\cdot 5=9996$

So the A.P. is considered up to $9996$

G.P.

General term: $g_n=4\cdot 2^{n-1}$

We want the largest 4-digit term: $4\cdot 2^{n-1}\le 9999$ $2^{n+1}\le 9999$

Now,

\qquad 2^{14}=16384>9999$$ Thus the largest 4-digit term in the G.P. is $$8192$$ This corresponds to $$4\cdot 2^{11}=8192$$ So the G.P. is considered up to $$8192$$ --- 5. **Find common terms of the two series** We need numbers common to both: - A.P.: $$11,16,21,26,31,36,\dots,9996$$ - G.P.: $$4,8,16,32,64,128,256,512,1024,2048,4096,8192$$ A common term must satisfy $$11+5k = 4\cdot 2^m$$ Now check the G.P. terms modulo $5$. Since A.P. terms are $$11,16,21,26,\dots$$ all are congruent to $$1 \pmod 5$$ So common terms must be G.P. terms congruent to $1 \pmod 5$. Now powers of $2$ modulo $5$ cycle as: $$2,4,3,1,2,4,3,1,\dots$$ Thus $$4\cdot 2^{n-1} \pmod 5$$ will cycle through: $$4\cdot 1=4, \ 4\cdot 2=3, \ 4\cdot 4=1, \ 4\cdot 3=2, \dots$$ So we can directly inspect the G.P. terms: $$4,8,16,32,64,128,256,512,1024,2048,4096,8192$$ Among these, - $$16 \equiv 1 \pmod 5$$ - $$256 \equiv 1 \pmod 5$$ - $$4096 \equiv 1 \pmod 5$$ Now verify they are in the A.P.: $$16=11+5(1)$$ $$256=11+5(49)$$ $$4096=11+5(817)$$ All are indeed terms of the A.P. The next such G.P. term would be $$65536$$ which is not 4-digit, so not allowed. Hence the common terms are: $$16,256,4096$$ Therefore, the number of common terms is $$\boxed{3}$$ --- 6. **Compare with stored correct answer** Stored correct answer: $3$ Our derived answer is also $3$, so it agrees.

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