Question

The sum of the 5th and the 7th term of an ap is 52 and the 10th term is 46 find the ap ​

Answer 1

Given that, sum of 5th & the 7th term of the Arithmetic Progression (AP) is 52. And also, 10th term is 46.

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Therefore,

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$:\implies\sf a + 9d = 46 \qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup\\\\\\:\implies\sf a + 4d + a + 6d = 52 \\\\\\:\implies\sf 2a + 10d = 52 \\\\\\:\implies\sf 2(a + 5d) = 52 \\\\\\:\implies\sf a + 5d = \dfrac{\cancel{52}}{\cancel{\:2}} \\\\\\:\implies\sf a + 5d = 26 \qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

⠀

$\dag\;{\underline{\frak{Now, \ Subtracting \;eq\:(2)\;from\;eq\:(1),}}}$

$:\implies\sf a + 9d = 46 \\\\\\:\implies\sf a + 5d = 26\\\\\\:\implies\sf 4d = 20\\\\\\:\implies\sf d = \dfrac{\cancel{20}}{\cancel{\:4}}\\\\\\:\implies\boxed{\frak{\pink{d = 5}}}$

$\therefore\;{\underline{\sf{Here, \ we \ get \ value \ of \ d \ is \; {\textsf{\textbf{5}}}.}}}$

⠀⠀⠀

☯ Now, Putting value of d in eq (2),

$:\implies\sf a + 5d = 26 \\\\\\:\implies\sf a + 5(5) = 26\qquad\qquad\bigg\lgroup\bf d = 5\bigg\rgroup\\\\\\:\implies\sf a + 25 = 26\\\\\\:\implies\sf a = 26 - 25\\\\\\:\implies\boxed{\frak{\pink{a = 1}}}$ ⠀

So now,

• a + d, 1 + 5 = 6
• $\sf a_2 + d$, 6 + 5 = 11
• $\sf a_3 + d$, 11 + 5 = 16

• 1, 6, 11, 16, 21, 26.... so on.

$\star\:\boxed{\textsf{This is required Arithmetic Progression}}$

Answer 2

■ QuésTion :-

➫ The sum of the 5th and the 7th term of an ap is 52 and the 10th term is 46 find the Ap.

■ AnsWer :-

➮ The Ap formed is 1 ,6 , 11 , 16 .............

■ Given :-

➢ The sum of the 5th and the 7th term of an ap is 52 and the 10th term is 46.

■ Formula used :-

$➤ \: \: \: \: \red{ \overline \red{\underline {\red{\boxed{ \green{\: \: \: \: \: \: \sf \: t_n = a + (n - 1)d \: \: \: \: \: \: \: \: }}}}}}$

■ SolutiOn :-

Let 'a' be the first term and 'd' be the the common difference , then ,

$\bigstar \sf \: a_5 = a + (5 - 1)d \: = a + 4d$

$\bigstar \sf \: a_7= a + (7 - 1)d \: = a + 6d$

Therefore,

$\: \: \: \: \: \implies \sf \: a_5 + a_7 = a + 4d + a + 6d = 52$

$\: \: \: \: \: \implies \sf \: 2 a + 10d = 52$

After dividing whole equation by 2 ,

$\: \: \: \: \: \implies \sf \: a + 5d = 26$

$\: \: \: \: \: \implies \sf \: a = 26 - 5d ............................. \color{yellow}\boxed{ \rm equation \: no.1}$

Similarly,

$\: \: \: \: \: \: \: \: \: \: \: \: \sf \: a_{10}= a + (10 - 1)d \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: = a + 9d \: = 46..........................(2)$

By substitution method, using eqⁿ 1 & 2 ,

26 - 5d = 9d - 46

➨ 9d - 5d = 46 - 26

➨ 4d = 20

➨d = 20/4

➨ d = 5

Now put the value of 'd' in equation 1,

$\: \: \: \: \: \sf \: a = 26 - 5 \times 5$

$\: \: \: \: \: ➩ \: \: \sf \: a = 26 - 25$

$\: \: \: \: \: ➩ \: \: \sf \: a = 1$

Now,

$\sf \: \: \: \: \: \: \: \: \: \: \: 1. \: \: \: \: a_2 = a_1 + d$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 + 5$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6$

$\sf \: \: \: \: \: \: \: \: \: \: \: 2. \: \: \: \: a_3 = a_2 + d$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6+ 5$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =11$

$\sf \: \: \: \: \: \: \: \: \: \: \: 3. \: \: \: \: a_4 = a_3 + d$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 11+ 5$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =16$

:. The Ap formed is 1,6,11,16..................

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➯ More formulae :-

$➤ \: \: \: \: \red{ \overline \red{\underline {\red{\boxed{ \green{\: \: \: \: \: \: \sf \: t_n = a + (n - 1)d \: \: \: \: \: \: \: \: }}}}}}$

$➤ \: \: \: \: \red{ \overline \red{\underline {\red{\boxed{ \green{\: \: \: \: \: \: \sf \: S_n = \frac{n}{2} [ \ \times 2a+ (n - 1)d ] \: \: \: \: \: \: \: \: }}}}}}\: \:$

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