1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

Let

Sn = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$

If 100 Sn = n, then n is equal to :
A
199
B
99
C
200
D
19

Explanation

nth term, Tn = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$

Tn = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$

$$ \Rightarrow $$ Tn = $${2 \over {n\left( {n + 1} \right)}}$$ = $$2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$

$$ \therefore $$ Sn = $$\sum {{T_n}} $$

= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$

= $$2\left( {1 - {1 \over n}} \right)$$

= $${{{2n} \over {n + 1}}}$$

Given that,

100 Sn = n

$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n

$$ \Rightarrow $$ n + 1 = 200

$$ \Rightarrow $$ n = 199
2
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
12 + 2.22 + 32 + 2.42 + 52 + 2.62 ...........
If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to
A
496
B
232
C
248
D
464

Explanation

Note :

Sum of square of first n odd terms

12 + 32 + 52 + . . . . .+ n2 = $${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$$

Given,

12 + 2. 22 + 32 + 2.42 + 52 + 2.62 + . . . . . .

A = Sum of first 20 terms

$$\therefore\,\,\,$$A = 12 + 2.22 + 32 + 242 + 52 + 2.62 + . . . . . .20 terms

Arrange those terms this way,

A = [12 + 32 + 52 + . . . . . 10 terms] + [ 2.22 + 2.42 + 2.62 + . . . . 10 terms]

A = [ 12 + 32 + 52 + . . . . 10 terms ] + 2.2 [ 12 + 22 + 32 + . . . .10 terms ]

A = $$ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$$

A = $$ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$$

A =70 $$ \times $$ 19 + 70 $$ \times $$ 44

A = 70 $$ \times $$ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [12+ 32 + 52 +. . . . 20 terms ] + [2.22 + 2.42 +. . . . . 20 terms ]

B = [12 + 32 + 52 + . . . . 20 terms] + 2.22 [12 + 22 + . . . 20 terms ]

B = $${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$$

B = 260 $$ \times $$ 41 + 560 $$ \times $$ 41

B = 41 $$ \times \,\,\,820$$

$$\therefore\,\,\,$$ B $$-$$ 2A = 41 $$ \times \,$$ 820 $$-$$ 2 $$ \times \,$$ 70 $$ \times \,$$ 63 = 24800

Given that B $$-$$ 2A = 100 $$\lambda $$

$$\therefore\,\,\,$$ 100 $$\lambda $$ = 24800

$$ \Rightarrow \,\,\,\lambda $$ = 248
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to
A
33
B
66
C
68
D
34

Explanation

a1, a2, a3 . . . a43 are in AP

So, a2 = a1 + d

a3 = a1 + 2d

.

.

.

a49 =a1 + 48d

Now given, $${a_9} + {a_{43}} = 66$$

$$ \Rightarrow \,\,\,\,$$ a1 + 8d + a1 + 42d = 66

$$ \Rightarrow \,\,\,\,$$ 2a1 + 50d = 66

$$ \Rightarrow \,\,\,\,$$ a1 + 25d = 33 . . . . . (1)

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416

$$ \Rightarrow \,\,\,\,$$ a1 + a5 + a9 + a13 +. . . . . 13 items = 416

$$ \Rightarrow \,\,\,\,$$ a1 + a1 + 4d + a1 + 8d + . . . . a1 + 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 4d +8d + 12d + . . . . . 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 4 (1+ 2 + 3 + . . . + 12) d = 416

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 24 $$ \times$$ 13d = 416

$$ \Rightarrow \,\,\,\,$$ a1 + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and $${a_1} = 8$$

$$\therefore\,\,\,$$ a1 = 8

a2 = 8 + 1 = 9

a3 = 8 + 2 = 10

.

.

.

a17 = 8 + 16 = 24

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (12 +22 + . . . . +242) $$-$$ (12 + 22 + . . . . .+ 72) = 140 m

$$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$$

$$ \Rightarrow \,\,\,\,\,$$ 490 $$-$$ 140 = 140 m

$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m

$$ \Rightarrow \,\,\,\,\,$$ m = 34
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

If x1, x2, . . ., xn and $${1 \over {{h_1}}}$$, $${1 \over {{h_2}}}$$, . . . , $${1 \over {{h_n}}}$$ are two A.P..s such that x3 = h2 = 8 and x8 = h7 = 20, then x5.h10 equals :
A
2560
B
2650
C
3200
D
1600

Explanation

Assume d1 is the common difference of A.P x1,x2 ..... xn

Given x3 = 8 and x8 = 20

$$ \therefore $$ x1 + 2d1 = 8 ..... (i)
and x1 + 7d1 = 20 ..... (ii)

Solving (i) and (ii) we get x1 = $$16 \over {15}$$ and d1 = $$12 \over {5}$$

Now let $$1 \over d_2$$ is the common difference of A.P $$1 \over h_1$$, $$1 \over h_2$$ ..... $$1 \over h_n$$

Given that,
h2 = 8 and h7 = 20

$$ \therefore $$ $$1 \over h_2$$ = $$1 \over 8$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$1 \over d_2$$ = $$1 \over 8$$ .... (iii)

and $$1 \over h_7$$ = $$1 \over 20$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$6 \over d_2$$ = $$1 \over 20$$ ... (iv)

Solving (iii) and (iv) we get
$$1 \over h_1$$ = $$28 \over 200$$ and $$1 \over d_2$$ = $$- {3 \over 200}$$
So, x5 = x1 + 4d1
= $$16 \over 5$$ + $$48 \over 5$$= $$64 \over 5$$ and
$$1 \over h_{10}$$ = $$1 \over h_1$$ + $$9 \over d_2$$
= $$28 \over 200$$ - $$27 \over 200$$ = $$1\over 200$$

$$ \therefore $$ x5 $$\times$$ h10 = $${64 \over 5} \times 200$$ = 2560

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