An arithmetic progression 5,12,19......has 50 terms. …
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Sia ? 6 years, 4 months ago
We have {tex}a = 5{/tex} and {tex}d = 12 - 5 = 7{/tex} and {tex}n = 50{/tex}
{tex}a_n = a + (n - 1)d{/tex}
{tex}\therefore{/tex} a50 = 5 + (50 -1 )7
= 5 + 49 {tex}\times{/tex} 7
= 5 + 343
= 348
Also the first term of the A.P of last 15 terms be a36
a36 = 5 + (36 - 1)7
= 5 + 35 {tex}\times{/tex} 7
= 5 + 245
= 250
Now, sum of last 15 terms
{tex}\therefore \quad S _ {15} = \frac { 15 } { 2 } [ 2 \times 250 + ( 15 - 1 ) 7 ]{/tex}
{tex}= \frac { 15 } { 2 } ( 500 + 14 \times 7 ){/tex}
{tex}= \frac { 15 } { 2 } ( 500 + 98 ){/tex}
{tex}= \frac { 15 } { 2 } \times 598{/tex}
= {tex}15 \times 299{/tex}
= 4485
Hence, sum of last 15 terms = 4485
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