Question

the first and the last termof an A.P.is 1 and 11 respectively if the sum of its terms is 36 then find thenumber of terms​

Answer 1

Given

• First term of the A.P = 1
• Last term of the A.P = 11
• Sum of its terms is 36.

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To find

• Number of terms in the A.P.

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Solution

• As we know

❍ First term = a

❍ Last term = l

❍ Number of terms = n

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• Using the formula

$\large{\boxed{\boxed{\bf{S_n = \dfrac{n}{2}\bigg\lgroup{a + l}\bigg\rgroup}}}}$

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• In this question

❍ Sum of the terms ($\sf{S_n}$) = 36

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$\tt\longmapsto{\dfrac{n}{2}[1 + 11] = 36}$

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$\tt\longmapsto{\dfrac{n}{2}[12] = 36}$

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$\tt\longmapsto{6n = 36}$

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$\tt\longmapsto{n = \dfrac{36}{6}}$

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$\bf\longmapsto{n = 6}$

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Hence,

• The number of terms in the given A.P is 6.

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Answer 2

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ɢɪᴠᴇɴ -

• First term of the A.P = 1
• Last term of the A.P = 11
• Sum of its terms is 36.

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ᴛᴏ ꜰɪɴᴅ -

• Number of terms in the A.P.

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ꜱᴏʟᴜᴛɪᴏɴ -

ᴀꜱ ᴡᴇ ᴋɴᴏᴡ

• First term = a
• Last term = l
• Number of terms = n

⠀

ᴜꜱɪɴɢ ᴛʜᴇ ꜰᴏʀᴍᴜʟᴀ

$\large{\boxed{\boxed{\bf{S_n = \dfrac{n}{2}\bigg\lgroup{a + l}\bigg\rgroup}}}}$

⠀

ɪɴ ᴛʜɪꜱ Qᴜᴇꜱᴛɪᴏɴ

• Sum of the terms $(\sf{S_n})$ = 36

⠀

$\tt\longmapsto{\dfrac{n}{2}[1 + 11] = 36}$

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$\tt\longmapsto{\dfrac{n}{2}[12] = 36}$

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$\tt\longmapsto{6n = 36}$

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$\tt\longmapsto{n = \dfrac{36}{6}}$

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$\bf\longmapsto{n = 6}$

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ʜᴇɴᴄᴇ,

ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏꜰ ᴛᴇʀᴍꜱ ɪɴ ᴛʜᴇ ɢɪᴠᴇɴ ᴀ.ᴘ ɪꜱ $\boxed{\bf{\red{6.}}}$

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