find the sum of the first ten terms of the parallelogram and the sum of the first ten terms of the sum of the two and third terms in a parallelogram of 22 and the first and fourth terms of the sum of 85
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Sum of the first n terms is given by
Sn=2n[2a+(n−1)d]
Putting n = 10, we get
S10=210[2a+(10−1)d]
210=5(2a+9d)
2a+9d=42 (1)
Sum of the last 15 terms is 2565
Sum of the first 50 terms - sum of the first 35 terms = 2565
S50−S35=2565
250[2a+(50−1)d]−235[2a+(35−1)d]=2565
25(2a+49d)−235(2a+34d)=2565
5(2a+49d)−27(2a+34d)=513
10a+245d−7a+119d=513
3a+126d=513
a+42d=171 (2)
Multiply the equation (2) with 2, we get
2a+84d=342 (3)
Subtracting (1) from (3)
d=4
Now, substituting the value of d in equation (1)
2a+9d=42
2a+9×4=42
2a=42−36
2a=6
a=3
So, the required AP is 3,7,11,15,19,23,27,31,35,39.
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