Question
The number of terms needed to get Sn = 0 in the A.P of
63, 60, 57. . . . .Solution
Here the sequence is 63, 60, 57.. . . . So a = 63, d = – 3 and n = ?, Sn= 0; Sn = (n/2) [2 x 63 + (n – 1)(-3)] = 0 ⇒ [126 – 3n + 3] = 0 ⇒ [129 – 3n] = 0 ⇒ n = 43 So in the given sequence 43 number of terms are required.
If (r + s + t) = 12 and (rs + st + tr) = 60, then find the value of (r² + s² + t²).
A, B and C are three points on a circle such that the angles subtended by the chord AB and AC at the centre O are 110° and 130°, respectively. Then th...
- If p² + q² + r² = 240 and pq – qr – rp = 64, then determine the value of (p + q – r) ².
Solve: 5 – 3x > 2x + 10
In Δ ABC, ∠ B= 68° and ∠ C 32°. Sides AB and AC are produced to points D and E, respectively. The bisectors of ∠ DBC and ∠...
If, 2x + y = 13, and 2xy = 12, and 2x > y, then find the value of 8x³ – y³.
- If x² - 2x + 1 = 0, then find the value of (x⁴ + x⁻⁴)(x + x⁻¹).
If √c + (1/√c) = 7, then find the value of c + (1/c).
If x + 1/x = 3 (x ≠ 0), find the value of x³ + 1/x³.
- If 2a + b = 15 and ab = 18, then the value of (4a² - b²) .
Relevant for Exams:
