Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
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Answered by
25
Step-by-step explanation:
a+d=14------(1)
a+2d=18------(2)
equation (1)-(2)
-d=-4 ya d=4
put d=4in equation (1)
a+4=14
a=14-4
a=10
Sn=51/2[ 2a+(n-1)d]
=51/2[2×10+(51-1)4]
=51/2[20+200]
=51/2×220
=51×110=5610
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