If 2p p+8 3p+1 are in a.p. then p is
WebIf p 1, p+3,3 p 1 are in A.P., then p is equal to:A. 4B. 2C. 4D. 2. Login. Study Materials. NCERT Solutions. NCERT Solutions For Class 12. ... AP Board. APSCERT Books; AP … WebIf 2p,p+10,3p+2 are in AP, the value of p = A 6 B 5 C 4 D 7 Hard Solution Verified by Toppr Correct option is A) Was this answer helpful? 0 0 Similar questions If A=1−2+3−4+5++99;B=2+4+6+8+..+100, then the value of A+B Medium View solution > …
If 2p p+8 3p+1 are in a.p. then p is
Did you know?
WebIf p + 1, 2P - 10, 1 - 4p 2 are three consecutive terms of an arithmetic progression, find the possible values of p A. -4, 2 B. -3, 4 11 C. - 4 11, 2 D. 5, -3 Correct Answer: Option C Explanation 2p - 10 = p + 1 + 1 − 4 P 2 2 (Arithmetic mean) = 2 (2p - 100 = p + 2 - 4P 2) = 4p - 20 = p + 2 - 4p 2 = 4p 2 + 3p - 22 = 0 = (p - 2) (4p + 11) = 0 http://www-math.mit.edu/~desole/781/hw8.pdf
WebSolutions for If p – 1, p + 3, 3p – 1 are in A.P., then p is equal to:a)-4b)4c)2d)-2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as … Web18.781 Homework 8 Due: 8th April 2014 Q1 (3.1(10)). Prove that if pis an odd prime, then x2 2(mod p) has solutions if and only if p 1(mod 8) or if p 7(mod 8). Proof. By Theorem 3:3, 2 p = ( 1)(p2 1)=8.As pis an odd prime, p= 8k+ rfor r2f1;3;5;7g. p2 1 = 64k2 +16kr+r2 1, so ( 1)(p2 1)=8 = ( 1)(r2 1)=8. (r2 1)=8 is even for r2f1;7gand odd for r2f3;5g. So 2 is a square …
WebG•'37516†€ G–ljÁ G–Ç3844 ¡ G˜g‹a G˜g39421‰À Gš Gš 40470‹` G›§Ž¡ G›§414‰Â G G @4 G G4189• 3 Gžç ¡ Gžç4290˜`4 G ‡ƒA ‡ ‡4ƒH @4 ‡¢'„á ‡¢'4485š4 ‡£Ç† ‡£Ç4599„á … Web2 mrt. 2024 · Best answer if a,b,c are in AP then 2b = a + c 2 b = a + c b − a = c − b b - a = c - b 2b = a + c 2 b = a + c p − 1, p + 3, 3p − 1 p - 1, p + 3, 3 p - 1 (p + 3) − (p − 1) = (3p − …
Webp2+8p+12=0 Two solutions were found : p = -2 p = -6 Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring p2+8p+12 The first term is, p2 its ... More Items
WebIf p is a prime of the form 4 k + 3, such that 2 p + 1 is also prime, it can be proven that 2 p + 1 2 p − 1. So you have a family of primes being a counterexample to the claim. It is … one day cricket highest centuryWebExplanation. 2p - 10 = p + 1 + 1 − 4 P 2 2 (Arithmetic mean) = 2 (2p - 100 = p + 2 - 4P 2) = 4p - 20 = p + 2 - 4p 2. = 4p 2 + 3p - 22 = 0. = (p - 2) (4p + 11) = 0. ∴ p = 2 or - 4 11. … one day cricket highest score player nameWeb0 Step 1: Calculate the (any variable)p value [NOTE:The definition of an arithmetic series is that the difference of two consecutive terms is always constant,] EXAMPLE: p, 5p - 8, 3p+8 (5p - 8) - (p) = (3p + 8) - (5p - 8) Step 2: Substitute the (any variable) p value in the given terms and simplify, to get the proper arithmetic series. one day cricket england v australiaWebIf you write p + 3 in place of p -2 then the answer is 4. p−1,p+3,3p−1 are in A.P. ∴2(p+3)=p−1+3p−1. ⇒2p+6=4p−2. ⇒2p=8. ∴p=4 one day cricket gameshttp://www-math.mit.edu/~desole/781/hw8.pdf one day cricket highest score listWeb2 aug. 2024 · The exam is scheduled to be held from 2nd May to 19th May 2024 and 13th June to 20th June 2024. This is for the 2024 examination cycle. Earlier, the SSC MTS 2024 Notification was released. SSC MTS 2024 were accepted till 24th Feb 2024 & this is also the closing day for eligibility consideration. one day cricket highest century listWeb(i) 2p – 1, 7 and 3p will be three consecutive terms of an A .P. ifIlnd term – Ist term =IIIrd term – II term⇒ 7 – (2p – 1) = 3p –7⇒ 7 – 2p + 1 = 3p – 7⇒ –2p – 3p = – 7 –8 ⇒ – 5p = – 15⇒ p = 3.(ii) 2p + 1, 13 and 5p – 3 will be three consecutive terms of an AP if13 – (2p + 1) = (5p – 3) – 13⇒ 13 – 2p – 1 = 5p – 3 – 13⇒ –2p – 5p ... one day cricket england v australia 2015