Question

find the 5th term from the end of an AP 10 7 4 ..... -62
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Answer 1

Given: The sequence of an AP series is 10, 7, 4 ... -62

Need to find: The 5th term from the end of the series.

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• a = 10, difference between each term in the series is 10 - 7, d = -3, Tn = -62.

$\\ \boxed{ \red { \sf{ {Formula : {T}_{n} = a + (n - 1)d }}}}$

We need to find n (number of terms) first,

$\\ \\ \implies \sf - 62 = 10 + (n - 1)-3 \\ \\ \implies \sf - 62 = 10 + 3n + 3\\ \\ \implies \sf - 62 = 13 + 3n\\ \\ \implies \sf 3n = 13 + 62\\ \\ \implies \sf 3n = 75\\ \\ \implies \sf n = \dfrac{75}{3} \\ \\ \implies \sf n = 25$

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∴ Total number of terms in the series is 25.

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• We are asked to find, The fifth term from the end of the series i.e 20th term.

Applying the same formula,

$\\ \\ \implies \sf {T}_{20} = 10 + (20 - 1) - 3 \\ \\ \implies \sf {T}_{20} = 10 - 60 + 3 \\ \\ \boxed{\implies \bf {T}_{20} = - 47}$

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$\therefore \underline{\sf The \: 5th \: term \: from \: the \: end \: of \: the \: series \: is \: -47}$

Answer 2

$\bull$ Giνєи :-

• The sequence of an Arithmetic progression is 10, 7, 4. . . - 62

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$\bull$ Tο ƒiи∂ :-

• 5th term of this A.P. from last

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$\bull$ Fοямυℓα υѕє∂ :-

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${ \color{tan} {\bigstar \:}} \color{teal}{\underline{\boxed{ \large \tt{ \: A_n = a + (n - 1)d }}}} \: \color{tan} \bigstar$

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Where,

• d = common difference between each term
• a = first term
• an = nth term
• n = number of terms

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✪ First of all we will find the common difference :-

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Therefore,

Common difference = 10 - 7 = 3

Now,

We will put the given values in the formula to get the number of terms.

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✪ Therefore, number of terms are :-

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$\sf : \implies \: - 62 = 10 + (n - 1) \times 3 \\ \\ \sf : \implies \: - 62 - 10 = 3n - 3 \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: n = \dfrac{ - 75}{ - 3} = 25 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Therefore, the number of terms in this A.P. is 25

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We need to find out the 5th last term of this A.P., that is 20th term of this A.P.

Now, again we will use the same formula to obtain that term. So,

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$\sf : \implies \: A_{20} = 10 + (20 - 1)3 \\ \\ \sf : \implies \: A_{20} =10 \: - \: 60 \: + \: 3 \: \\ \\ \sf : \implies \: A_{20} =13 \: - \: 60 \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: A_{20} = - \: 47 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the 20th term of this A.P. is $\color{maroon} \boxed{ \bf \: - 47}$