if the sum of first terms of an AP whose second and third terms are 14 and 18 respectively
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Answered by
1
Answer:
Step-by-step explanation:
I assume you meant first n terms
a+d=14
a+2d=18
On solving
d=4
a=10
Sn=n/2*(20+(n-1)*4)
Answered by
3
a2 = 14 and a3 = 18
Common difference = a3 - a2 = 18 - 14 = 4 = d
Now
a2 = a+d=14
a+4=14
a = 10
Now, sum of 51 terms
={51(2a+(50)d)}/2
={51(20+200)}/2
={51×220}/2
=51×110=5610
Therefore sum of 51 terms is 5610
Common difference = a3 - a2 = 18 - 14 = 4 = d
Now
a2 = a+d=14
a+4=14
a = 10
Now, sum of 51 terms
={51(2a+(50)d)}/2
={51(20+200)}/2
={51×220}/2
=51×110=5610
Therefore sum of 51 terms is 5610
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