The sum of first n terms of an AP is n÷2×(3n+13) find it's 38th term.
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sₙ =n÷2[3n+13]
sₙ-₁ =(n-1)÷2[3n-3+13]
tₙ= sₙ-sₙ-₁
=>(n)÷2[3n+13] -(n-1)÷2[3n-3+13]
=>N/2[3 n+13] - (n-1)/2[3 n+10]
=>3n²/2+13n/2 - [3 n(n-1)/2 + 5(n-1)
=>3n²/2+13n/2 -3n²/2+3n/2 - 5n+5
=>13n/2+3n/2-5n +5
=>(16n - 10n)/3 + 5= tₙ
=>t₃₈= [16*38 - 10*38]/3+5
=>[608 - 380]/3+5
=>(228)/3 +5
=>76+5
t₃₈=81
hope its help you :-)
sₙ-₁ =(n-1)÷2[3n-3+13]
tₙ= sₙ-sₙ-₁
=>(n)÷2[3n+13] -(n-1)÷2[3n-3+13]
=>N/2[3 n+13] - (n-1)/2[3 n+10]
=>3n²/2+13n/2 - [3 n(n-1)/2 + 5(n-1)
=>3n²/2+13n/2 -3n²/2+3n/2 - 5n+5
=>13n/2+3n/2-5n +5
=>(16n - 10n)/3 + 5= tₙ
=>t₃₈= [16*38 - 10*38]/3+5
=>[608 - 380]/3+5
=>(228)/3 +5
=>76+5
t₃₈=81
hope its help you :-)
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