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sum of first 14 term of an ap is 1505 and its first term is 10 find its 25 term

Answer» Ans is 135
S14=1505 &A=10 we know that Sn=n/2[2a+(n-1)d ] here S14=14/2[2×10 +1(14-1)d ] 1505 = 7(20+13d)1505/7=20+13d215=20+13d215-20=13d195=13dd=195/13d=15Now A25=a+24d = 10+24×15 = 10+360 A25=370 is the answer
S14=15051505=n/2 (2a+(n-1)d 1505=14/2 (2 (10)+(14-1)d1505=7 (20+13)d 1505=7 (43)d 1505=301 (d)5=doA25=a+24d 10+24×5 10+120 130
Answer is 120


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