Question

Find the 16th term of an A.P., whose 1st term is 15 and common difference is -2.

Answer 1

Answer:

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Step-by-step explanation:

a= 15 and d = -2

a16=?

a + 15d

15 + 15 × -2

15-30

-15

therefore a16 = -15

Answer 2

Answer:

• 16th term of an A.P = -15

Explanation:

$\underline{\sf{\bigstar\:Given\:information}}$

$\sf Find\:the\:16^{th}\:term\:of\:an\:A.P\:whose\:1^{st}\:term$ $\sf is\:15\:and\:common\:difference\:is\:-2.$

• $\sf First\:term\:(a) = \bf{15}$
• $\sf Common\:difference\:(d) = \bf{-2}$
• $\sf a_{16} =\:?$
• $\sf Number\:of\:terms\:(n) = \bf{16}$

$\underline{\sf{Using\:formula,}}$

$\qquad\bf{\dag}\:{\underline{\boxed{\bf{\pink{a_{n} = a + (n - 1)d}}}}}$

$\underline{\sf{Putting\:all\:values,}}$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 + (16 - 1)-2\qquad\quad-\:(1)$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 + (15)-2$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 + (15\:\times\:-2)$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 + (-30)$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 - 30$

$\\ \twoheadrightarrow\:\bf \Big|\red{a_{16} = -15}\Big|$

• $\small{\underline{\bf{Hence,\:16^{th}\:term\:of\:an\:A.P\:is\:-15.}}}$

$\underline{\sf{\bigstar\:Verification}}$

$\\ \twoheadrightarrow\:\sf a_{16} = 15 + (16 - 1)-2\qquad-\:From\:(1)$

$\\ \underline{\sf{Putting\:value\:of\:a_{16}\:in\:above\:eq^{n},}}$

$\\ \twoheadrightarrow\:\sf -15 = 15 + (16 - 1)-2$

$\\ \twoheadrightarrow\:\sf -15 = 15 + (15)-2$

$\\ \twoheadrightarrow\:\sf -15 = 15 + (15\:\times\:-2)$

$\\ \twoheadrightarrow\:\sf -15 = 15 + (-30)$

$\\ \twoheadrightarrow\:\sf -15 = 15 - 30$

$\\ \twoheadrightarrow\:\sf -15 = -15$

$\\ \twoheadrightarrow\:\bf \Big|\purple{LHS = RHS}\Big|$

• $\underline{\bf{Hence,\:Verified\:\checkmark}}$

$\\ {\underline{\sf{\bigstar\:Knowledge\:BoosteR}}}$

• $\underline{\bf{What\:is\:an\:A.P?}}$

An arithmetic progression (A.P) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

• $\underline{\bf{What\:is\:common\:difference?}}$

The fixed number which is added to preceding term to get next term is called common difference. Note :: Common difference can be zero, negative or positive.

$\qquad{\boxed{\underline{\underline{\bf{\bigstar\:Formulae\:related\:to\:A.P\:\bigstar}}}}}$

$\:$

◐ $\underline{\sf{n^{th}\:term\:of\:an\:A.P}}$

$\\ \mapsto\:\boxed{\bf{a_{n} = a + (n - 1)d}}$

$\:$

◐ $\underline{\sf{Sum\:of\:n\:terms\:of\:an\:A.P}}$

$\\ \mapsto\:\boxed{\bf{S_{n} = \dfrac{n}{2}\Big[2a + (n - 1)d\Big]}}$

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