Question
An arithmetic sequence consists of 21 terms. The sum of
the three middle terms is 360, while the sum of the last three terms amounts to 1170. Determine the eighth term of this sequence.Solution
ATQ,
Let first term and common dierence of the arithmetic progression be ‘a’ and ‘d’ respectively. So, middle three terms of the arithmetic progression are 10th , 11th and 12th term. So, a + 9d + a + 10d + a + 11d = 360 Or, 3a + 30d = 360 And, a + 18d + a + 19d + a + 20d = 1170 Or, 3a + 57d = 1170 Or, 360 – 30d + 57d = 1170 Or, 27d = 810 Or, d = 30 And, 3a = 360 – 30 × 300 = -540 Or, a = -180 So, 8th term = a + 7d = -180 + 30 × 7 = 30Â
124, 132, 118, 140, 108, 148
- Find the wrong number in the given number series.
5, 10, 20, 35, 80, 160 26, 38, 60, 110, 206, 398
60, 63, 71, 86, 125, 145
12, 18, 14, 17, 16, 15, 18, 15
8 10.4 6.8 10.2 7.6 10
21, 32, 45, 60, 75, 96
Find the wrong number in the given number series.
276, 291, 308, 329, 341
768Â Â Â 2304Â Â Â 288Â Â Â 864Â Â Â 106Â Â Â 324
275, 50, 165, 33, 99, 19.8, 59.4