The first term,common difference and last term of an AP are 12,68 and 352 respectively. Find the sum of all terms of this AP.
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a = 12 ,d= 68 ,an= 352
an = a+ (n-1)d = 352
an = 12+ (n-1)68=352
an= 12+ 68n-68= 352
an= 68n-56=352
an= 68n=352+56
an= 68n= 408
an = n = 408/68
n=6
Sn = n/2[2a+(n-1)d]
Sn= 6/2[2(12)+ (6-1)68]
Sn= 3[24+ 340]
Sn= 3(364)
Sn=1092
hope it helps you.
please mark it as brainliest.
an = a+ (n-1)d = 352
an = 12+ (n-1)68=352
an= 12+ 68n-68= 352
an= 68n-56=352
an= 68n=352+56
an= 68n= 408
an = n = 408/68
n=6
Sn = n/2[2a+(n-1)d]
Sn= 6/2[2(12)+ (6-1)68]
Sn= 3[24+ 340]
Sn= 3(364)
Sn=1092
hope it helps you.
please mark it as brainliest.
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