Question

What is the first term of A.P. if common difference is -5 and 18th term is -68.​

Answer 1

Answer:

find a18= using formula

an=a+ (n-1)d

Answer 2

GivEn:

• Common difference, d = -5

• $\sf 18^{th}$ term of AP, $\sf a_{18}$ = -68

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To find:

• First term of AP?

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SoluTion:

$\dag\;{\underline{\frak{Using\; Arithmetic\; progression\;formula,}}}$

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$\star\;{\boxed{\sf{\purple{ a_{n} = a + (n - 1)d}}}}$

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${\frak{Here}} \begin{cases} & \text{d = -5 } \\ & \text{ \sf a_{18} = -68 } \\ & \text{n = 18} \end{cases}$

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$\dag\;{\underline{\frak{Putting\;values,}}}$

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$:\implies\sf a_{18} = a + (18 - 1)(-5)$

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$:\implies\sf - 68 = a + (18 - 1)(-5)$

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$:\implies\sf - 68 = a + 17 \times -5$

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$:\implies\sf - 68 = a - 85$

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$:\implies\sf a = -68 + 85$

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$:\implies{\underline{\boxed{\sf{\purple{a = 17}}}}}\;\bigstar$

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$\therefore$ Hence, first term of AP is 17.

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$\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$

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$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; \red{a_n + (n - 1)d}$

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$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \; \purple{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

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$\sf (iii)\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; \pink{ \dfrac{n}{2}(a + l)}$