The fifth term of an A.P. is thrice the
second term and 12th term exceeds
twice the 6th term by 1. Find the
16th term.
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Given that fifth term of an A.P. is thrice the second term.
∴a5=3a2
As we know that
an=a+(n−1)d
∴a+4d=3(a+d)
⇒2a−d=0
⇒d=2a⟶(1)
Also given that twelfth term exceeds twice the 6th term by 1.
∴a12=2a6+1
⇒a+11d=2(a+5d)+1
⇒a+11d=2a+10d+1
⇒a−d+1=0⟶(2)
From eqn(1)&(2), we have
⇒a−2a+1=0
⇒a=1
∴d=2a=2×1=2
∴a16=a+15d=1+15×2=1+30=31
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