Mathematics - Sequence & Series Question with Solution | TestHub
Match each item in List I to all applicable entries in List II.
List I | List II |
|---|---|
(A) For an A.P. a,b,10,c,d, find a+b+c+d. | (P) 10 |
(B) The sides of a right triangle form a positive G.P. with shortest side 2. Its hypotenuse is 1+√5. Find 1²+5². | (Q) 20 |
(C) The first three terms of a convergent G.P. sum to 70; multiplying the extremes by 4 and the middle by 5 gives an A.P. Find its first term. | (R) 26 |
(D) A parallelogram has diagonals 4 m and 6 m, with angle 60° between them. Its perimeter is 2(√m+√n) metres, for positive integers m,n. Find m+n. | (S) 40 |
Options:
Answer:
Solution:
(A) Opposite terms of the A.P. sum to 20, so a+b+c+d=40. (B) Write the sides as 2,2r,2r², r>1. Pythagoras gives r⁴=1+r², hence the hypotenuse 2r²=1+√5 and 1²+5²=26. (C) 4a+4ar²=10ar gives r=1/2 or 2. Convergence selects 1/2; a(1+1/2+1/4)=70 gives a=40. (D) Using half-diagonals 2 and 3, the side lengths are √(4+9±12 cos60°)=√19 and √7. Thus m+n=26. Therefore A-S; B-R; C-S; D-R.
