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Sequences and Series question

2022 · 27 Jul · Shift 2 · Q27
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  5. /2022 · 27 Jul · Shift 2 · Q27

Sequences and Series question

2022 · 27 Jul · Shift 2 · Q27

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 9825\frac{98}{25}2598​. Then the sum of the first 21 terms of an AP, whose first term is 10ar,nth 10\mathrm{a r}, \mathrm{n}^{\text {th }}10ar,nth  term is an\mathrm{a}_{\mathrm{n}}an​ and the common difference is 10ar210 \mathrm{ar}^{2}10ar2, is equal to :
  1. A
    21 a1121 \,\mathrm{a}_{11}21a11​
  2. B
    22 a1122 \,\mathrm{a}_{11}22a11​
  3. C
    15 a1615 \,\mathrm{a}_{16}15a16​
  4. D
    14 a1614 \,\mathrm{a}_{16}14a16​
View written solutionFree

Correct answer: A

  1. Use the GP conditions

For an infinite G.P. with first term aaa and common ratio rrr,

a1−r=5⇒a=5(1−r).\frac{a}{1-r}=5 \quad \Rightarrow \quad a=5(1-r).1−ra​=5⇒a=5(1−r).

Also, the sum of the first five terms is

S5=a(1−r5)1−r=9825.S_5=\frac{a(1-r^5)}{1-r}=\frac{98}{25}.S5​=1−ra(1−r5)​=2598​.

Using a1−r=5\dfrac{a}{1-r}=51−ra​=5, we get

5(1−r5)=9825.5(1-r^5)=\frac{98}{25}.5(1−r5)=2598​.

So,

1−r5=98125⇒r5=1−98125=27125=(35)5.1-r^5=\frac{98}{125} \quad \Rightarrow \quad r^5=1-\frac{98}{125}=\frac{27}{125}=\left(\frac35\right)^5.1−r5=12598​⇒r5=1−12598​=12527​=(53​)5.

Hence,

r=35.r=\frac35.r=53​.

Then

a=5(1−35)=5⋅25=2.a=5\left(1-\frac35\right)=5\cdot \frac25=2.a=5(1−53​)=5⋅52​=2.

So,

a=2,r=35.a=2,\qquad r=\frac35.a=2,r=53​.
  1. Form the AP

The AP has:

  • first term =10ar=10ar=10ar
  • common difference =10ar2=10ar^2=10ar2

Compute these:

10ar=10⋅2⋅35=12,10ar=10\cdot 2\cdot \frac35=12,10ar=10⋅2⋅53​=12, 10ar2=10⋅2⋅(35)2=20⋅925=365.10ar^2=10\cdot 2\cdot \left(\frac35\right)^2=20\cdot \frac{9}{25}=\frac{36}{5}.10ar2=10⋅2⋅(53​)2=20⋅259​=536​.

Thus the AP is:

12,  12+365,  …12,\; 12+\frac{36}{5},\; \dots12,12+536​,…

with first term A=12A=12A=12 and common difference d=365d=\frac{36}{5}d=536​.

  1. Find the sum of first 21 terms

Sum of first 212121 terms of an AP:

S21=212[2A+(21−1)d].S_{21}=\frac{21}{2}\left[2A+(21-1)d\right].S21​=221​[2A+(21−1)d].

Substitute A=12A=12A=12, d=365d=\frac{36}{5}d=536​:

S21=212[24+20⋅365]=212[24+144]=212⋅168=21⋅84=1764.S_{21}=\frac{21}{2}\left[24+20\cdot \frac{36}{5}\right] =\frac{21}{2}\left[24+144\right] =\frac{21}{2}\cdot 168 =21\cdot 84 =1764.S21​=221​[24+20⋅536​]=221​[24+144]=221​⋅168=21⋅84=1764.
  1. Compute a11a_{11}a11​ and a16a_{16}a16​

For the AP,

an=A+(n−1)d=12+(n−1)365.a_n=A+(n-1)d=12+(n-1)\frac{36}{5}.an​=A+(n−1)d=12+(n−1)536​.

So,

a11=12+10⋅365=12+72=84.a_{11}=12+10\cdot \frac{36}{5}=12+72=84.a11​=12+10⋅536​=12+72=84.

Hence,

21a11=21⋅84=1764.21a_{11}=21\cdot 84=1764.21a11​=21⋅84=1764.

This matches S21S_{21}S21​.

Also,

a16=12+15⋅365=12+108=120.a_{16}=12+15\cdot \frac{36}{5}=12+108=120.a16​=12+15⋅536​=12+108=120.

Then,

15a16=1800,14a16=1680,15a_{16}=1800, \qquad 14a_{16}=1680,15a16​=1800,14a16​=1680,

which do not match.

  1. Check options
  • A: 21a11=176421a_{11}=176421a11​=1764 ✅
  • B: 22a11=184822a_{11}=184822a11​=1848 ❌
  • C: 15a16=180015a_{16}=180015a16​=1800 ❌
  • D: 14a16=168014a_{16}=168014a16​=1680 ❌

Therefore, the correct option is

A.\boxed{A}.A​.
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