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
More Algebra Questions
- What will come in place of the question mark (?) in the following questions?
300−40% of 200=? - Simplify the following expression:
16 + [17 - (8 + 11) + 6 - 3] ÷ 0.2 Simplify the following expression.
(3-3 × 3 + 3 ÷ 3 + 3 × 5) × 2 of 5 + (2 + 2 ÷ 2 + 2 × 2 - 2)
2/9 of 5/8 of 3/25 of ? = 40
115 ÷ 23 + 12 × 6 = ? + 16 - 35
√? = 32% of 900 + 48% of 50
 Â
5/13 × 104 + 1(2/9) × 198 = 133 + ?
Simplify the following expressions and choose the correct option.
40% of 360 + 25% of 248 - 30
(11/12) × (18/22) × (4/3) + 3 = ?2
[(15)³ × (8)²] ÷ (90 × 6) = ?²
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