an AP consists of 37 terms. the sum of the three middle most terms is 225.and the sum of the last three terms is 429. find the AP
utkbkkh:
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S3=429
Middle term=a18+a19+a20=225------(1)
Also,
a35+a36+a37=429---------(2)
You can solve these two equations to get the value of a1 and then you can easily get the AP. So lets solve,
In equation 1,
We can write a18+a19+a20 as,
a+17d+a+18d+a+19d=225
3a+57d=225-----(3)
Also,
a+34d+a+35d+a+36d=429
3a+105d=429-------(4)
Subtracting eq-4 to 3, we get
48d=204
So,
d=4.25
Now,putting the value of d in eq 3 we get
3a+105*4.25=429
3a+446.25=429
3a=-17.25
a=-5.75
AP=-5.75,1.25,5.5
Hope you like my answer and brainliest it
Middle term=a18+a19+a20=225------(1)
Also,
a35+a36+a37=429---------(2)
You can solve these two equations to get the value of a1 and then you can easily get the AP. So lets solve,
In equation 1,
We can write a18+a19+a20 as,
a+17d+a+18d+a+19d=225
3a+57d=225-----(3)
Also,
a+34d+a+35d+a+36d=429
3a+105d=429-------(4)
Subtracting eq-4 to 3, we get
48d=204
So,
d=4.25
Now,putting the value of d in eq 3 we get
3a+105*4.25=429
3a+446.25=429
3a=-17.25
a=-5.75
AP=-5.75,1.25,5.5
Hope you like my answer and brainliest it
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