1) if x, y, z are in A P such that a1+a5+a10+a15+a20+a24=300, then a1+a2+a3+ +a24 is - Maths - Arithmetic Progressions

Question

1) if x, y, z are in A P such that a1+a5+a10+a15+a20+a24=300,
then a1+a2+a3+ +a24 is equal to?

Ajanta Trivedi · Accepted Answer

if the given terms are in AP let the first term be $a_{1} = a$ let the common difference be d therefore the nth term is $a_{n} = a + (n - 1) d$ the given sum is $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 300 a + (a_{5} + a_{10} + a_{15} + a_{20} + a_{25}) + (a_{24} - a_{25}) = 300 a + \frac{5}{2} [a_{5} + a_{25}] - d = 300 [s u m o f 5 t e r m s]$ where first term is a5 and last term is a25 since difference between two consecutive terms is equal to the common difference therefore the above equation is: $a + \frac{5}{2} [a + 4 d + a + 24 d] - d = 300 a + \frac{5}{2} [2 a + 28 d] - d = 300 a + 5 a + 70 d - d = 300 6 a + 69 d = 300 3 (2 a + 23 d) = 300 2 a + 23 d = \frac{300}{3} = 100 (1)$ now sum of the first 24 terms is $S_{24} = \frac{24}{2} [a_{1} + a_{24}] S_{24} = 12 [a + a + 23 d] S_{24} = 12 (2 a + 23 d)$ substituting the value from (1): $S_{24} = 12 * 100 = 1200$ thus the sum of the first 24 terms is 1200

1) if x, y, z are in A.P. such that a1+a5+a10+a15+a20+a24=300, then a1+a2+a3+....+a24 is equal to?