WebIn an AP, if S 5 + S 7 = 167 and S 10 = 235, then find the AP, Where S n denotes the sum of its first n terms. 4564 Views Switch Flag Bookmark Advertisement The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers. Answer WebJun 19, 2024 · The sum of first 5 terms of an ap and the sum of first 7 terms of an ap is 167 if sum of first 10 terms of the ap is 235 find the sum of its first 20 terms Advertisement pradu4321 is waiting for your help. …
In an ap S5 + S7 is equal to 167 and S10 is 235 find the AP where …
WebIn an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms. Solution: Question 25. The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference Solution: Question 26. WebFeb 13, 2016 · In an AP, if S5 + S7 = 167 and S10 = 235, then find the A P , where Sn denotes the sum of its first n term - Maths - Arithmetic Progressions. NCERT Solutions; Board Paper Solutions; ... 2 by 6, we get 12 a + 54 d = 282 .... 3 Subtracting 1 from 3, we get 54 d - 31 d = 282-167 ⇒ 23 d = 115 ⇒ d = ... ipsos offices
In an AP. It is given that S5 + S7 = 167 and S10 = 235 , then find the
WebLet a and d be the first term and the common difference of the AP, respectively Chapter Chosen. Arithmetic Progressions Book Chosen. Mathematics Subject Chosen. Mathematics ... In an AP, if S 5 + S 7 = 167 and S 10 = 235, then find the AP, Where S n denotes the sum of its first n terms. Let a and d be the first term and the common difference of ... WebIn an AP, if S 5 + S 7 = 167 and S 10 = 235, then find the AP, Where S n denotes the sum of its first n terms. Let a and d be the first term and the common difference of the AP, respectively. 4564 Views. Switch; Flag; Bookmark; Advertisement . 2. In the given figure, PQ is a chord of a circle with centre O and PT is tangent. WebJan 24, 2024 · Step-by-step explanation: Answer We know that sum of n terms sn = n/2 (2a + (n - 1) * d) Given s5 + s7 = 167. = 5/2 (2a + (5 - 1) * d) + 7/2 (2a + (7 - 1) d) = 167 = 5/2 (2a + 4d) + 7/2 (2a + 6d) = 167 = 5 (a + 2d) + 7 (a + 3d) = 167 = 5a + 10d + 7a + 21d = 167 = 12a + 31d = 167 ---------- (1) Given that s10 = 235 10/2 (2a + (10 - 1) * d) = 235 ipsos new york address