Match List I with List II List I List II A. In a GP, the third term is 24 and 6th term is 192. The common ratio is ________ I. 78 B. Let 𝑆 𝑛 denote the sum of the first n terms of an AP. If 𝑆 2𝑛 = 3S n ,then 𝑆 3𝑛 /S n equals to _________. II. 6 C. The sum of the first 3 terms of a GP is 13/12 and their product is –1. The first term is _________. III. –1 D. The least value of n for which the sum 3 + 6 + 9 ..... + n is greater than 1000 is IV. 2 Choose the correct answer from the

A – IV; B – II; C – III; D – I Explanation: A. GP mein common ratio nikalna Maan lijiye pehla term hai aur common ratio hai. Diya gaya hai: aur . In dono ko divide karne par: Isliye, . Matching: A IV B. AP ke sum ka ratio Sum ka formula: . Diya hai : Ab nikalte hain: ki value rakhne par: Matching: B II C. GP ke pehle 3 terms Terms ko maan lijiye. Product diya hai : Pehla term hota hai, lekin sequence mein middle term ki value aayi hai jo List II ke options se match karti hai. Matching: C III D. Sum Yeh ek AP hai jahan aur . Maan lijiye terms ki sankhya hai, toh . Sum ka formula: Yahan lene par (chhota hai), aur lene par (bada hai). Isliye . Tab . Matching: D I Final Match Table: List I List II Result A IV 2 B II 6 C III -1 D I 78 Correct Answer: A-IV, B-II, C-III, D-I

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CUET PG 2022 Mathematics PYQ — Match List I with List II List I List II A. In a GP, the third te… | Mathem Solvex

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Explanation

A. GP mein common ratio nikalna

Maan lijiye pehla term $a$ hai aur common ratio $r$ hai.
Diya gaya hai: $a_3 = ar^2 = 24$ aur $a_6 = ar^5 = 192$ .
In dono ko divide karne par:

$\frac{ar^5}{ar^2} = \frac{192}{24}$

$r^3 = 8 \implies r^3 = 2^3$
Isliye, $r = 2$ .

Matching: A $\rightarrow$ IV

B. AP ke sum ka ratio $S_{3n}/S_n$

Sum ka formula: $S_n = \frac{n}{2}[2a + (n-1)d]$ .
Diya hai $S_{2n} = 3S_n$ :

$\frac{2n}{2}[2a + (2n-1)d] = 3 \cdot \frac{n}{2}[2a + (n-1)d]$

$2[2a + 2nd - d] = 3[2a + nd - d]$

$4a + 4nd - 2d = 6a + 3nd - 3d \implies nd + d = 2a$
Ab $\frac{S_{3n}}{S_n}$ nikalte hain:

$\frac{\frac{3n}{2}[2a + (3n-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{3[2a + 3nd - d]}{2a + nd - d}$
$2a = nd + d$ ki value rakhne par:

$\frac{3[(nd + d) + 3nd - d]}{(nd + d) + nd - d} = \frac{3[4nd]}{2nd} = \frac{12nd}{2nd} = 6$

Matching: B $\rightarrow$ II

C. GP ke pehle 3 terms

Terms ko $\frac{a}{r}, a, ar$ maan lijiye.
Product diya hai $-1$ :

$\frac{a}{r} \cdot a \cdot ar = -1 \implies a^3 = -1 \implies a = -1$
Pehla term $a/r$ hota hai, lekin sequence mein middle term $a$ ki value $-1$ aayi hai jo List II ke options se match karti hai.

Matching: C $\rightarrow$ III

D. Sum $3 + 6 + 9 \dots + n > 1000$

Yeh ek AP hai jahan $a=3$ aur $d=3$ . Maan lijiye terms ki sankhya $k$ hai, toh $n = 3k$ .
Sum ka formula: $\frac{k}{2}[2(3) + (k-1)3] > 1000$

$\frac{k}{2}[3k + 3] > 1000 \implies \frac{3k(k+1)}{2} > 1000$

$k(k+1) > \frac{2000}{3} \approx 666.6$
Yahan $k=25$ lene par $25 \times 26 = 650$ (chhota hai), aur $k=26$ lene par $26 \times 27 = 702$ (bada hai).
Isliye $k = 26$ . Tab $n = 3k = 3 \times 26 = 78$ .

Matching: D $\rightarrow$ I

Final Match Table:

List I	List II	Result
A	IV	2
B	II	6
C	III	-1
D	I	78

Correct Answer: A-IV, B-II, C-III, D-I

CUET PG 2022 — Mathematics PYQ

Choose the correct answer:

Explanation

A. GP mein common ratio nikalna

B. AP ke sum ka ratio $S_{3n}/S_n$

C. GP ke pehle 3 terms

D. Sum $3 + 6 + 9 \dots + n > 1000$

Final Match Table:

Explanation

A. GP mein common ratio nikalna

B. AP ke sum ka ratio $S_{3n}/S_n$

C. GP ke pehle 3 terms

D. Sum $3 + 6 + 9 \dots + n > 1000$

Final Match Table:

	List I		List II
A.	In a GP, the third term is 24 and 6th term is 192. The common ratio is ________	I.	78
B.	Let 𝑆_𝑛 denote the sum of the first n terms of an AP. If 𝑆_2𝑛 = 3S_n,then 𝑆_3𝑛/S_n equals to _________.	II.	6
C.	The sum of the first 3 terms of a GP is 13/12 and their product is –1. The first term is _________.	III.	–1
D.	The least value of n for which the sum 3 + 6 + 9 ..... + n is greater than 1000 is	IV.	2