Question

If a 1 = 34, a2 = 32 find the sum of terms upto 15 term ​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

•a = 34

•d = a₂ - a₁

= 32 - 34

= - 2

•n = 15

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

•The sum of the numbers upto 15 term

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\:\:\:\:\dag\bf{\underline \green{Putting\:the\:values:-}}$

$\bf S_n = \frac{n}{2}[2a +(n-1)d]\\\\\longrightarrow\sf S_{15}=\frac{15}{2} [ 2(34)+(15-1)(-2)]\\\\\longrightarrow \sf S_{15}= \frac{15}{2}[68+ 14 (-2)]\\\\\longrightarrow \sf S_{15}= \frac{15}{2}[68-28]\\\\\longrightarrow \sf S_{15}= \frac{15}{2} × 40\\\\\longrightarrow \sf S_{15}= 15 × 20\\\\\green{\boxed{\bf \therefore S_{15}= 300}}$

$\:\:\:\:\dag\bf{\underline{\underline \green{Hence:-}}}$

• The sum of 15 terms of the given set of numbers is 300.

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$\;\;\underline{\textbf{\textsf{ Know More :-}}}$

•nth term of an AP formulas:-

$\\\\\sf 1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf 2) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf 3) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 4) \: Difference \: of \: two \: terms = (m - n)d$

$\\\sf 5) \:Middle\: term\: of\: a\: finite\: AP$

$\sf (i) \:\: If \: n \: is \: odd = \frac{n + 1}{2}\:th\:term$

$\sf (ii) \:\: If \: n \: is \: even = \frac{n}{2} \:th \: term \: and \: ( \frac{n}{2} + 1)th \: term$

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•Sum Formulas :-

$\sf 1) \: Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf 2) \: Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf 3) \: Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$

Answer 2

Answer:

300

Step-by-step explanation:

here a1=34;  d=-2;   n=15

S=n/2[2a+(n-1)d]

S=15/2[2(34)+(15-1)(-2)

S=15/2[68-28]

S=15/2*40

S=300

expecting a sweet thanks....