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The common difference of an A.P. is given byd = S 2 – 2S 1 where S 2 is the sum of first two termsand S 1 is the sum of first term or the first term.If sum of n terms S n is given then general termT n = S n – S n–1 , where S n–1 is sum of (n – 1) terms ofA.P.If for an A.P. sum of p term is q and sum of q termsis p, then sum of (p + q) terms is {–(p + q)}.If for an A.P., sum of p term is equal to sum of qterms, then sum of (p + q) terms is zero.If the p th term of an A.P. is 1/q and q th term is 1/p,then sum of pq terms is given by S pq = 21 (pq + 1).Sum of n A.M.'s between a and b is equal to n timesthe single A.M. between a and b.⎛ a + b ⎞i.e. A 1 + A 2 + A 3 + ...... + A n = n ⎜ ⎟ .⎝ 2 ⎠If T k and T p of any G.P. are given, then formula for1⎛ T n kn⎞ − ⎛ Tp⎞ p−kobtaining T n is⎜T⎟ = ⎜ ⎟⎝ k ⎠ T⎝ k ⎠Product of n G.M.'s between a and b is equal to n thpower of single geometric mean between a and bi.e. G 1 G 2 G 3 .... G n = ( ab ) n .The product of n geometric means between a and 1/ais 1.⎛ ⎞If n G.M.'s inserted between a and b then r = ⎜ ⎟⎝ a ⎠11b n+1Quadratic Equations and Inequations :An equation of degree n has n roots, real orimaginary.If f(α) = 0 and f´(α) = 0, then α is a repeated root ofthe quadratic equation f(x) = 0 and f(x) = a(x – α) 2 .In fact α = –b/2a.If α is repeated common root of two quadraticequations f(x) = 0 and φ(x) = 0, then α is also acommon root of the equations f´(x) = 0 and φ´(x) = 0.In the equation ax 2 + bx + = 0 [a, b, c ∈R], ifa + b + c = 0 then the roots are 1, c/a and if a – b + c= 0, then the roots are –1 and – c/a.If one root of the quadratic equation ax 2 + bx + c = 0is equal to the n th power of the other, thenn( ) 1ac n+ 1 + ( n 1c) n 1a + + b = 0.If one root is k times the other root of the quadratic2equation ax 2 (k + 1) b 2+ bx + c = 0, then = .k acIf an equation has only one change of sign, it has one+ve root and no more.Permutations and Combinations :n C 0 = n C n = 1, n C 1 = nn C r + n C r–1 = n+1 C rn C x = n C y ⇔ x = y or x + y = nn. n-1 C r–1 = (n – r + 1) n C r–1If n is even then the greatest value of n C r is n C n/2 .If n is odd then the greatest value of n C r isnC n − 1.2n C r = rn . n–1 C r–1 .nC 1n +or2Number of selection of zero or more things out of ndifferent things is, n C 0 + n C 1 + b C 2 + ... + n C n = 2 n .n C 0 + n C 2 + n C 4 + .... = n C 1 + n C 3 + n C 5 + .... = 2 n–1 .Gap method : Suppose 5 moles A, B, C, D, E arearranged in a row as × A × B × C × D × E ×. Therewill be six gaps between these five. Four in betweenand two at either end. Now if three females P, Q, Rare to be arranged so that no two are together weshall use gap method i.e., arrange them in betweenthese 6 gaps. Hence the answer will be 6 P 3 .Together : Suppose we have to arrange 5 persons ina row which can be done in 5! = 120 ways. But if twoparticular persons are to be together always, then wetie these two particular persons with a string. Thuswe have 5 – 2 + 1 (1 corresponding to these twotogether) = 3 + 1 = 4 units, which can be arranged in4! ways. Now we loosen the string and these twoparticular can be arranged in 2! ways. Thus totalarrangements = 24 × 2 = 48.If we are given n different digits (a , a 2 , a 3 ..... a n ) thensum of the digits in the unit place of all numbersformed without repetition is (n – 1)!(a 1 + a 2 + a 3 + ....+ a n ). Sum of the total numbers in this case can beobtatined by applying the formula (n – 1)!(a 1 + a 2 +a 3 + ..... + a n ). (1111 ......... n times).Binomial Theorem & Mathematical Induction :The number of terms in the expansion of (x + y) n are(n + 1).If the coefficients of p th , q th terms in the expansion of(1 – x) n are equal, then p + q = n + 2.For finding the greatest term in the expansion of(x + y) n . we rewrite the expansion in this formn(x + y) n = x n ⎡ y⎢ 1 ⎤ +x⎥ . Greatest term in (x + y) n = x n .⎣ ⎦n⎛ y ⎞Greatest term in ⎜1 + ⎟ .⎝ x ⎠There are infinite number of terms in the expansionof (1 +x) n , when n is a negative integer or a fraction.The number of term in the expansion of(x 1 + x 2 + .... + x 2 ) n = n+r-1 C r–1 .XtraEdge for IIT-JEE 49 APRIL 2010