Question

4.

Find the 15th term of the arithmetic sequence whose common difference is d=9 and whose first term is a, = 2. 8 Х 5 ?

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8

Answer 1

4 2 For a anithmetic sequence, a, ( first term) = a (Commen difference) = 9 15th term of the arithmetic sequence is alt (15-1)d 2 + (15-1). 9 2 + 19.9 tis = 2 + 126 128 Hence Aus. 128 Let a, bethe first term & d be the commen difference of a arithmatic sequence Given that asy =-233 27 ait (89-1) 22-233 => alt 88 2 = -233 aq=7 a +(91) a=7 alt 8a 7

Subtracting ① from 6 we get (a1 +882) - (9,+83) = -233-7 ar = -240 From D alt 8(-3) 27 a - 24-7 a12 7+2 4 2 31 + a33 ait (33-1)d 314 (33) (-3) = 31+ 328 (-3) = 31 96 65 2 Hence ayz = -65] # Since - 908 - 408 3,0-3, is an arithmatic sequence whose affirst term) d (common difference) = 0-3= -2 nith term of the seanence be tn alt n-1) a = - 408 3+ (0-1) (-3) = - 408 (-3)(n-1 -408-3 = - 411 n-1- 2137 13741 = 138. tetal rumbers of term of the sequence is 138 n =

2. Sn = 2 [zept (n-1) a] 133 [2.3+(138-1).(-2)] 69 [6 - 411 69X (-405) - 27945 Hence 3+0+(-3) + .+ (-408) = - 27945 # For i=1, -zi+9= -2.149 = 7 For i =2, -21+9= -2.2+9 = 65 For 123 - 26 ta = -2.3tq = 3 Fur i=108 -zita= -2x108+9=-207 Po - Thure the sequence 7,S, 3, sequence whye a = 7 5-7=-2 - 207 is an artimatic 8. Slog 2 n-108 (total numbers of term) [291 +6108-1) a 54 [ 2x7 +(108–1).(-2)] 254 [ 14 + 107*(-2) = 54 [19-214 59 X (-20) - 1080 Hone 108 I (-2i+9) = -10800 il

7) and verter of (4,9) and (09, -3) be the center of on ellipse. the major asus of the ellipse is vertically Therefore the canation of the ellipse is of the so form (x-4) (8-4) + av since 21 (arybr) length of major any minor anis is a 2b = 4 6 = 2 br=4. 24 verters of the elipse & one (4, Ata) since (4,-3) is a vertex often without loss of generated ata=-3 a=-3-4=-7 =7 7 av = (-7) 2 = 49. From O 4. 4 8) (x-4) (19) 49 which is the required evention of lipse. since (-3,5) betere verton of the Prabalar the graph, the earation of the para bela in of the form (y-p) 40(x-4)! (06) - vester => Y-5) = 4(x+3) and by

since the porabala passes through the point (-2,1) a from (1-5) = 40 (-2+3) > (na)² = 49.1 qa=16 From (3-5) = 16 (x+3) which is the reanirea avertin porabala, tim of the