Web14 mei 2024 · Correct answer is C. P3 Let first term = a and Common difference = d ∴ According to the question, Sn = n2 p Sn = n/2 (2a + (n–1) d) = n2p 2a + (n–1) d = 2np………………. (1) And Sm = m2p Sm= m/2 (2a + (m–1) d) = m2p 2mp = (2a + (m–1) d)………………. (2) Subtracting 2 from 1 2a + (n–1) d – 2a – (m–1) d = 2np – 2mp d … WebC OL OR A DO S P R I N G S NEWSPAPER T' rn arr scares fear to speak for the n *n and ike UWC. ti«(y fire slaves tch> ’n > » t \ m the nght i »ik two fir three'."—J. R. Lowed W E A T H E R F O R E C A S T P I K E S P E A K R E G IO N — Scattered anew flu m e * , h igh e r m ountain* today, otherw ise fa ir through Sunday.
If the sum of first m terms of an AP is n and the sum of first n …
WebIf Sm=Sn for some A.P, then prove that Sm+n=0 Arithmetic Progression 624 views Dec 11, 2024 18 Dislike Share VipraMinds (Rahul Sir) 44K subscribers If Sm=Sn for some A.P, then prove... Web7 okt. 2024 · knjroopa Answer: Step-by-step explanation: Given ratio of the sum of m terms sm/sn=m^2/n^2 show that ratio of mth term am/an=2m-1/2n-1 We know that sum of n terms of an A.P. Sn = n/2 (2a + (n – 1) d) Similarly sum of m terms will be Sm = m/2 (2a + (m – 1)d) Given ratio Sm / Sn = m^2 / n^2 m/2 (2a + (m – 1)d) / n/2 (2a + (n – 1)d) = m^2 / n^2 draw a mechanism for the following reaction
In an A. P., Sm : Sn = m^2 : n^2 The ratio of p^2 th term to q^2 …
Web13 aug. 2011 · Sn= (a1+an)n/2 = (a1+a1+nd-d)n/2 =n (a1-d/2)+dn²/2 =an²+bn 其中,a= (a1-d/2),b=d/2 这么设是为了表示的方便而已。 所以有, Sn=an²+bn=m Sm=am²+bm=n 故Sm-Sn= (am²+bm)- (an²+bn)=a (m-n) (m+n)+b (m-n)=n-m 得知a (m+n)+b=-1 则Sm+n=a (n+m)²+b (n+m) = (n+m) [a (n+m)+b] =- (m+n) 以上~ 101 评论 gotoxyz 2024-06-04 · TA … Web26 jul. 2024 · Best answer Let the first term of the AP be a and the common difference be d Given: Sm = m2p and Sn = n2p To prove: Sp = p3 According to the problem (m - n)d = 2p (m - n) Now m is not equal to n So d = 2p Substituting in 1st equation we get Hence proved. ← Prev Question Next Question → Find MCQs & Mock Test Free JEE Main Mock Test Web1 sep. 2024 · Given; In an A.P, Sm = n and Sn = m, also m > n To Find; the sum of the first ( m-n ) terms. Solution; It is given that In an A.P, Sm = n and Sn = m, also m > n Sn=n2 [2a+ (n−1)d] =m Sm=m2 [2a+ (m−1)d]=n Sn−Sm=n2 [2a+ (n−1)d]−m2 [2a+ (m−1)d] 2 (m−n)=2a (n−m)+ [ (n2−m2)− (n−m)]d−2 (n−m)= (n−m) [2a+ { (n+m)−1}d] Divide (n-m) … employee great job performance