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If the 20th term of an AP=560 and 30th term of AP=840 then what is the sum of 5th term and 40th term of the series.
a) 1450 b) 1560 c) 1260 d) 1340
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- sorry..made a blunder
20th term T20=560 & 30th term T30=840
nth term=1st term + (n-1)difference
so T20= T1 + (n-1)d =T1 +19d
& T30=T1 +(n-1)d =T1 +29d
solving both, d=28 & T1=28
so T5+T40= T1 +4d +T1 +39d
=2T1 +33d
=2*28+43*28
=1260 - 12 years agoHelpfull: Yes(28) No(0)
- Ans:1260 20th term=560 30th term=840 diff is 280 then 10th term is 280 so fifth term is 140 and 40th term=30th term=10th term=840+280=1120 so sum of 5th and 40th terms=1120+140=1260
- 12 years agoHelpfull: Yes(26) No(0)
- apply t=a +n-1 *d and by this u wil get 2 equations and then using them u will a=28 and d=28 and finding 5 and 40th term adding them is 140+1120=1260
- 12 years agoHelpfull: Yes(13) No(0)
- let a be 1st term and d is common difference of the AP
20th term=a+19d=560
30th term=a+29d=840
from the difference of above 2 equations 10d=280
so d=28
a=560-19d
we need 40th term=a+39d
=560-19d+39d
=560+20d
=560+(20*28)
=560+560
1120... - 12 years agoHelpfull: Yes(8) No(13)
- 20th term T20=560 & 30th term T30=840
nth term=1st term + (n-1)difference
so T20= T1 + (n-1)d =T1 +19d
& T30=T1 +(n-1)d =T1 +29d
solving both, d=28 & T1=28
so T5+T40= T1 +4d +T1 +39d
=2T1 +33d
=2*28+33*28
=980 - 12 years agoHelpfull: Yes(0) No(19)
- padmi_phoenix@yahoo.com
- 12 years agoHelpfull: Yes(0) No(2)
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