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In an ap s5+s7 167

WebIn an AP, S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms urgently need this answer!!!! - Maths - Arithmetic Progressions WebIn an AP if S5+S7 =167 and S10 =235 then find S6. Class 10 cbse AP extra questions RR Maths Mantra 5 subscribers Subscribe 37 views 10 months ago #cbseclass10 …

In an A.P., S5 + S7 = 167 and S10 = 235, then find the A.P., where …

WebIn an A.P., S5 + S7 = 167 and S10 = 235, then find the A.P., where Sn denotes the sum of first n terms. Class 11 >> Applied Mathematics >> Sequences and series >> Arithmetic … WebIn an AP, it is given that S 5 + S 7 = 167 a n d S 10 = 235, then find the AP, where S n denotes the sum of its first n terms. Q. In an AP, if S 5 + S 7 = 167 and S 10 = 235 , then find the … highway express iom https://spumabali.com

In an AP, S5 + S7 = 167 and S10 = 235, then find the AP, where Sn ...

WebJul 9, 2024 · closed Jul 10, 2024 by Anaswara In an AP. It is given that S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms. arithmetic progression class-10 1 Answer +2 votes answered Jul 9, 2024 by Dheeya (31.0k points) selected Jul 10, 2024 by Anaswara Best answer WebIn an AP. It is given that S5 + S7 = 167 and S10 = 235 , then find the AP, where Sn denotes the sum of its first n terms In an AP. It is given that S5 + S7 = 167 and S10 = 235 , then find the AP, where Sn denotes the sum of its first n terms Arithmetic Progression CBSE Class 10 Maths RS Aggarwal In an AP. 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 = ... highway exit design

In an AP,it is given that S5+S7 =167 and S10=235 then …

Category:The sum of first 5 terms of an ap and the sum of first 7 …

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In an ap s5+s7 167

In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, Where …

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. … WebJan 19, 2024 · 14K views 3 years ago Arithmetic Progressions In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms. Show more The …

In an ap s5+s7 167

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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 WebJan 2, 2024 · thesarcasticsoul Answer: S5+S7=5/2(2a+4d) +7/2(2a+6d)=167 =5a+10d+7a+21d=167 =12a+31d=167 [let this be eqn 1] similarly, from S10=235, we get 2a+9d=47 {let this be equation 2} by solving eqn 1 and 2, we get a=1 and d=5 therefore the ap is 1,6,11........ Step-by-step explanation: Advertisement Advertisement New questions in …

WebIn an A.P., if S5 + S7 = 167 and S 10= 235, then find the A.P., where Sn denotes the sum of its first n terms. Advertisement Remove all ads Solution S S and S S 5 + S 7 = 167 and S 10 = … WebNov 6, 2024 · In an AP,it is given that S5+S7 =167 and S10=235 then find the AP,where Sn denotes the sum of its first n terms. See answers Advertisement BigSmoke Here your …

WebOct 15, 2024 · S5+S7=167 therefore 2a+10d=167 and a+9d=235 from these 2 equations we get d=12.625 therefore the ap starts with 40.75 and continue s with a difference d=12.625 olf they have given sum of terms not nth term, so the formula you're using is wrong...……... Find Math textbook solutions? See all Class Class 12 Class Class 11 Class Class 10 Class … WebApr 12, 2024 · In an AP, If S5 + S7 = 167 and S10 = 235 then find the AP, where Sn - YouTube In this video i solved important questions of class 10 maths for term 2,the question is as followingIn an AP,...

Web>> In an A.P., S5 + S7 = 167 and S10 = 2. ... Verified by Toppr. Was this answer helpful? 0. 0. Similar questions. If the 3rd and the 9th terms of an AP are 4 and -8 respectively. which term of this AP IS ZERO ? Medium. View solution > If the sum of 7 terms of an A.P is 4 9 and that of 1 7 terms is 2 8 9, find the sum of n terms.

WebNov 30, 2024 · In an AP, if S5 +S7 =167 and S10 =235, then find the AP, where Sn denotes the sum of its first n terms. [2015] ...[2M] Corporate Off. The world’s only live instant tutoring platform. Become a tutor About us Student login Tutor login. Login. Student Tutor. Filo instant Ask button for chrome browser. ... small streamer partnershipsWebMar 19, 2024 · In an AP, if S5+S7 =167 and S10=235, then find the AP, where Sn denotes the sum of its first n terms. 15. Let a be the flrst term and d be the common तilt of: the given … small streamers connectWebIn an AP S5+S7=167 and S10=235then find the AP where Sn; The points A47Bp3 and C73 are the vertices of a right a; Find the relation between x and y if the points Axy B-5; The 14th term of an AP is twice its 8th If its 6th te highway express lines trucksWebIn 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. highway express transport bramptonWebThe sum of the sum of first five terms of an AP and the sum of the first seven terms of the same AP is 167. ... Consider an A.P. whose first term and the common difference are a and d respectively. According to the question: S5 + S7 = 167 (Given) \Rightarrow \frac{5}{2}[2 a+(5-1) d]+\frac{7}{2}[2 a+(7-1) d]=167\\ \Rightarrow 5\{2 a+4 d\}+7\{2 ... highway exit sign clip art 485WebIt is given that S5 + S7 = 167 and S10 = 235 , then find the AP, where Sn denotes the sum of its first n terms. Arithmetic Progression CBSE Class 10 Maths RS Aggarwal. In an AP. … small streamer gamesWebJan 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 highway etymology