1st series is 1, 3, 5, 7 , .... 120 terms
Last term = a + (120 -1) x d = 1 + 119 x 2 = 239
Hence, 1, 3, 5, 7,..., 239 is the first series.

2nd series 3, 6, 9, .., 80 term.
Last term = 3 + (80 - 1) x 3 = 240
Hence, 3, 6 , 9 ....240 is the last term

From above two series, the first common term is 3 and next is 9 and so on, Hence , the series of common terms is 3, 9, 15,.. last term.

Correct Option: B

1st series is 1, 3, 5, 7 , .... 120 terms
Last term = a + (120 -1) x d = 1 + 119 x 2 = 239
Hence, 1, 3, 5, 7,..., 239 is the first series.

2nd series 3, 6, 9, .., 80 term.
Last term = 3 + (80 - 1) x 3 = 240
Hence, 3, 6 , 9 ....240 is the last term

From above two series, the first common term is 3 and next is 9 and so on, Hence , the series of common terms is 3, 9, 15,.. last term.

Last common term is 237 (occurring in both).
Thus, the number of terms is this series which will give the number of common terms between the two series
(l - a)/(d + 1) = (237 - 3)/(6 + 1) = 40

Here, the 1st AP is (1 + 6 + 11 + ...)
and 2nd AP is (4 + 5 + 6 + ...)

Correct Option: A

Here, the 1st AP is (1 + 6 + 11 + ...)
and 2nd AP is (4 + 5 + 6 + ...)

1st AP = (1 + 6 + 11 + ..)
Here , common difference = 5 and the number of terms = 100
∴ Sum of series = S₁ = n/2[2a + (n - 1)d]
= 100/2 [2 x 1 + (100 - 1) x 5 ]
= 50[2 + 99 x 5 ]
= 50 x 497
= 24850

2nd AP = (4 +5 + 6 + ...)
Here, common difference = 1
and number of terms = 100
S₂ = 100/2[2 x 4 + 99 x 1]
= 50 x 107 = 5350
∴ Sum of the given series
= S₁ + S₂ = 24850 + 5350
= 30200

Let the man first save ₹ P in the first year.
Then, the given sequence is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, S_n = 145000

S_n = n/2[2a + (n - 1)d]
⇒ 145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]

Correct Option: B

Let the man first save ₹ P in the first year.
Then, the given sequence is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, S_n = 145000

S_n = n/2[2a + (n - 1)d]
⇒ 145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]
⇒ 2P + 18000 = 145000/5 = 29000
⇒ 2P = 29000 - 18000
⇒ 2P = 11000
∴ P = ₹ 5500
So, the man save ₹ 5500 in the first year .

Let the four parts be (a - 3d), (a - d), (a +d), (a + 3d).
From question,
(a -3d) + (a - d) + (a + b) + (a + 3d) = 20
⇒ a = 5
Also, [(a - 3d) (a +3d)] / [(a - d) (a + d)] = 2/3

Correct Option: B

Let the four parts be (a - 3d), (a - d), (a +d), (a + 3d).
From question,
(a -3d) + (a - d) + (a + b) + (a + 3d) = 20
⇒ a = 5
Also, [(a - 3d) (a +3d)] / [(a - d) (a + d)] = 2/3
⇒ [a² - 9d²] / [a² - d²] = 2/3
⇒ [5² - 9d²] / [5² - d²] = 2/3
⇒ d = 1
∴ a = 5, d = 1
So, the four parts are 2, 4, 6, 8

Using formula ,

1³ + 2³ + 3³ +...+ n³ =		n(n + 1)		²
		2

Correct Option: D

Using formula ,

1³ + 2³ + 3³ +...+ n³ =		n(n + 1)		²
		2

1³ + 2³ + 3³ + ..... + 10³ =		10 × 11		²	= 55²
		2