Question

Find the sum of all numbers from 50 to 250 which are divisible by 6 and find t13

Answer 1

Sum of all numbers = 4950
t13 = 126

Answer 2

$54, 60, 66, 72,\cdot\cdot\cdot ,246 \:are \\ numbers \: which \:are \:divisible \:by \: 6 \: and \\and \: it \: is \:an \:A.P$

$First \: term (a) = 54$

$Common \: differnce (d) = a_{2} - a_{1} \\= 60 - 54 \\= 6$

$n^{th} \: term ( t_{n}) = 246$

$\boxed{ \pink { n^{th} \:term (t_{n}) = a+(n-1)d }}$

$\implies a + (n-1)d = 246$

$\implies 54 + (n-1)6 = 246$

/* On Dividing bothsides by 6, we get */

$\implies 9 + n - 1 = 41$

$\implies 8 + n = 41$

$\implies n = 41 - 8$

$\implies n = 33$

$\boxed { \pink { Sum \:of \:n \:terms (S_{n}) = \frac{n}{2} ( a + t_{n} ) }}$

$Here, a = 54, n = 33 \:and \: t_{n} = 246$

$i )Sum \:of \: 33 \: terms (S_{33}) \\= \frac{33}{2}(54+246) \\= \frac{33}{2}\times 300 \\= 33 \times 150 \\= 4950$

$ii )Now, t_{13} = a + 12d \\= 54 + 12 \times 6 \\= 54 + 72 \\= 126$

Therefore.,

$\red{ Sum \:of \:all \: numbers \:from \:50 \:to }\\\red{ 250 \: which \:are \: divisible \:by \:6 } \green {= 4950 }$

$and \: \red{ t_{13} } \green {= 126 }$

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