Question
In an A.P 41st term is 50, then the sum of 81 terms of
that A.P is:Solution
Here 41st terms is 50 So, a 41 = 50 ⇒ [a +(41 – 1)d] = 50 ⇒ a + 40d = 50 …………….(i) S81 = ?                 ⇒ Sn = (n/2) [2 x a + (n – 1)d] = 0 ⇒ S81 = (81/2)[2a + (81 – 1)d] ⇒ 81[a + 40d]……..eq(ii) Now putting the value from eq(i) in eq(ii) then ⇒ S81 = 81[50] = 4050
(11/12) × (18/22) × (4/3) + 3 = ?2
13/3 – (23/6) = ? – (22/9)
15 × 35 ÷7 + 60% of 300 =?
What will come in the place of question mark (?) in the given expression?
? = (27 × 13) – 26% of (412 – 92 )
25% of (?) + (1/4)of 5600 = 2500 – 20% of 1940Â
Determine the value of 'p' in following expression:
720 ÷ 9 + 640 ÷ 16 - p = √121 X 5 + 6²- 7?2 = (1035 ÷ 23) × (1080 ÷ 24)Â
[564 + 32 of 18 × 9 ÷ 12 + 162 ] ÷ 4 = ?
√14400 x √8100 - (60)² = ? + (80)²
What will come in the place of question mark (?) in the given expression?
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