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Sequences and Series question

2019 · 10 Jan · Shift 2 · Q36
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Sequences and Series question

2019 · 10 Jan · Shift 2 · Q36

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let a1, a2, a3, ..... a10 be in G.P. with ai > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k ∈\in∈ N (the set of natural numbers) for which ∣log⁡e a1ra2klog⁡e a2ra3klog⁡e a3ra4klog⁡e a4ra5klog⁡e a5ra6klog⁡e a6ra7klog⁡e a7ra8klog⁡e a8ra9klog⁡e a9ra10k∣\left| {\begin{matrix} {{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \\ {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \\ {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \\ \end{matrix} } \right|​loge​a1​ra2​kloge​a4​ra5​kloge​a7​ra8​k​loge​a2​ra3​kloge​a5​ra6​kloge​a8​ra9​k​loge​a3​ra4​kloge​a6​ra7​kloge​a9​ra10​k​​ === 0. Then the number of elements in S, is -
  1. A
    10
  2. B
    4
  3. C
    2
  4. D
    infinitely many
View written solutionFree

Correct answer: D

  1. Write the GP in logarithmic form

Let the G.P. be an=a qn−1,a>0, q>0a_n = a\,q^{n-1}, \qquad a>0,\ q>0an​=aqn−1,a>0, q>0 since all ai>0a_i>0ai​>0.

Define xn=ln⁡an.x_n = \ln a_n.xn​=lnan​. Then xn=ln⁡a+(n−1)ln⁡q,x_n = \ln a + (n-1)\ln q,xn​=lna+(n−1)lnq, so xnx_nxn​ is an arithmetic progression: xn=A+(n−1)Dx_n = A + (n-1)Dxn​=A+(n−1)D for some constants A=ln⁡aA=\ln aA=lna and D=ln⁡qD=\ln qD=lnq.

  1. Simplify each matrix entry

The (i,j)(i,j)(i,j)-type entries are of the form ln⁡(amram+1k)=rln⁡am+kln⁡am+1=rxm+kxm+1.\ln\left(a_m^r a_{m+1}^k\right)= r\ln a_m + k\ln a_{m+1} = r x_m + k x_{m+1}.ln(amr​am+1k​)=rlnam​+klnam+1​=rxm​+kxm+1​.

So the matrix is

r x_1 + k x_2 & r x_2 + k x_3 & r x_3 + k x_4 \\ r x_4 + k x_5 & r x_5 + k x_6 & r x_6 + k x_7 \\ r x_7 + k x_8 & r x_8 + k x_9 & r x_9 + k x_{10} \end{pmatrix}.$$ Since $x_n=A+(n-1)D$, $$r x_m + k x_{m+1} = r\big(A+(m-1)D\big)+k\big(A+mD\big).$$ This equals $$ (r+k)A + \big(r(m-1)+km\big)D = (r+k)A + \big((r+k)m-r\big)D.$$ Hence each entry is a linear function of $m$. --- 3. **Observe the pattern of the matrix** Let $$y_m = r x_m + k x_{m+1}.$$ Then $y_m$ is also an arithmetic progression in $m$. The matrix becomes $$M=\begin{pmatrix} y_1 & y_2 & y_3\\ y_4 & y_5 & y_6\\ y_7 & y_8 & y_9 \end{pmatrix}.$$ Because $y_m$ is an A.P., say $$y_m = P + (m-1)Q,$$ we get $$\text{Row}_1 = (P,\ P+Q,\ P+2Q),$$ $$\text{Row}_2 = (P+3Q,\ P+4Q,\ P+5Q),$$ $$\text{Row}_3 = (P+6Q,\ P+7Q,\ P+8Q).$$ Now, $$\text{Row}_2-\text{Row}_1 = (3Q,3Q,3Q),$$ $$\text{Row}_3-\text{Row}_2 = (3Q,3Q,3Q).$$ Therefore, $$\text{Row}_3 - 2\text{Row}_2 + \text{Row}_1 = 0.$$ So the three rows are linearly dependent. Hence, $$\det(M)=0$$ for **every** natural number pair $(r,k)$. --- 4. **Conclude the set $S$** Since the determinant is zero for all $$r,k\in \mathbb N,$$ every pair $(r,k)$ belongs to $S$. Thus $S$ has infinitely many elements. --- 5. **Check options** - A: $10$ — incorrect - B: $4$ — incorrect - C: $2$ — incorrect - D: infinitely many — correct Therefore, the number of elements in $S$ is $$\boxed{\text{infinitely many}}.$$
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