the sum of first seven terms of an arithemetic sequence is 119 and the sum of fitst 11th term is 275 what is 4th term
Answers
Answer:
Given:
Sum of first 7 terms of an AP (S7) = 119
Sum of first 20 terms of the AP (S20) = 860.
We know that,
Sum of first n terms of an AP = n/2 * [ 2a + (n - 1)d ]
Hence,
→ S(7) = 7/2 * [ 2a + (7 - 1)d ]
→ S(7) = 7/2 * [ 2a + 6d ]
→ 7/2 * [ 2 ( a + 3d) ] = 119
→ 7 ( a + 3d ) = 119
→ a + 3d = 17
We know,
nth term of an AP (an) = a + (n - 1)d
Hence,
a(4) = a + (4 - 1)d
→ a(4) = a + 3d
→ a(4) = 17 - (a)
Similarly,
S(20) = 20/2 * [ 2a + (20 - 1)d ]
→ 860 = 10 (2a + 19d)
→ 860/10 = 2a + 19d
→ 2a + 19d = 86
→ (a + 3d) + a + 16d = 86
→ (17) + (a + 16d) = 86
→ a + 16d = 86 - 17
→ a(17) = 69 - (b)
[ a(17) = a + (17 - 1)d ]
Hence, the 4th and 17th terms of the given AP are 17 , 69.
Step-by-step explanation:
Answer:
4 th term is 17 is the answer