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Sia ? 6 years, 6 months ago
Here Sum of the n terms of an Arithmetic progression, Sn = 5n2 + 3n
S1 = 5 {tex}\times{/tex} 12 + 3 {tex}\times{/tex} 1 = 8 = t1 ....(i)
S2 = 5 {tex}\times{/tex} 22 + 3 {tex}\times{/tex} 2 = 26 = t1 + t2 ...(ii)
S3 = 5 {tex}\times{/tex} 32 + 3 {tex}\times{/tex} 3 = 54 = t1 + t2 + t3 ...(iii)
From (i), (ii) and (iii),
t1 = 8, t2 = 18, t3 = 28
So Common difference, d = 18 - 8 = 10 and first term a=8.
Now tm = 168 (given)
We know that {tex}{T_m}=a+(m-1)d{/tex}
{tex}\Rightarrow{/tex} a + (m - 1)d = 168
{tex}\Rightarrow{/tex} 8 + (m - 1) {tex}\times{/tex} 10 = 168
{tex}\Rightarrow{/tex} (m - 1) {tex}\times{/tex} 10 = 160
{tex}\implies{/tex} m-1 ={tex}\frac{160}{10}{/tex}
{tex}\Rightarrow{/tex} m - 1 = 16
Therefore, m = 17
So, t20 = a + (20 - 1)d
= 8 + 19 {tex}\times{/tex} 10
= 8 + 190
= 198
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Abhishek Patwa 7 years, 3 months ago
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