Question

The sum of the first 10 terms of an arithmetic sequence is 350 and the sum of the first 15 terms is 100.Write the algebraic expression for the sequence?

Answer 1

GIVEN:

• The sum of the first 10 terms of an arithmetic sequence is 350
• The sum of the first 15 terms is 100.

TO FIND:

• What is arithmetic expression of sequence ?

SOLUTION:

CASE:- 1)

The sum of the first 10 terms of an arithmetic sequence is 350

We know that the formula for finding the sum of AP is:-

$\large{\boxed{\bf{\star \: {S}_{n} = \dfrac{n}{2} [2a +(n-1)d] \: \star}}}$

According to question:-

$\bf{\longmapsto {S}_{10} = 350}$

$\sf{\longmapsto 350= \cancel\dfrac{10}{2}[2a +(10-1) d] }$

$\sf{\longmapsto 350 = 5[2a + 9d]}$

$\sf{\longmapsto \cancel\dfrac{350}{5} = 2a + 9d }$

$\bf{\longmapsto 70 = 2a + 9d.....1) }$

$\sf{\longmapsto \dfrac{70-9d}{2} = a}$

CASE:- 2)

The sum of the first 15 terms is 100.

$\bf{\longmapsto {S}_{15} = 100}$

$\sf{\longmapsto 100= \cancel\dfrac{15}{2}[2a +(15-1) d] }$

$\sf{\longmapsto 100 \times 2 = 15[2a + 14d]}$

$\sf{\longmapsto \cancel\dfrac{200}{15} = 2a + 14d }$

$\sf{\longmapsto \dfrac{40}{3} = 2a + 14d}$

$\sf{\longmapsto 40 = 3(2a + 14d) }$

$\sf{\longmapsto 40 = 6a + 42d }$

Take '2' as common from both sides

$\bf{\longmapsto 20 = 3a + 21d....2) }$

Put the value of a in equation 2)

$\sf{\longmapsto 20 = 3 \bigg(\dfrac{70-9d}{2} \bigg) + 21d}$

$\sf{\longmapsto 20 = \dfrac{210 - 27d}{2} +21d }$

$\sf{\longmapsto 20 = \dfrac{210-27d+42d}{2} }$

$\sf{\longmapsto 40 = 210 + 15d}$

$\sf{\longmapsto 40-210 = 15d }$

$\sf{\longmapsto -170 = 15d }$

$\sf{\longmapsto \cancel\dfrac{-170}{15} = d }$

$\bf{\longmapsto \dfrac{-34}{3} = d}$

Put the value of d in equation 2)

$\sf{\longmapsto 20 = 3a + 21\bigg( \dfrac{-34}{3} \bigg)}$

$\sf{\longmapsto 20 = 3a - \dfrac{-714}{3}}$

$\sf{\longmapsto 20 = \dfrac{9a-714}{3}}$

$\sf{\longmapsto 60 = 9a -714}$

$\sf{\longmapsto 60+714 = 9a}$

$\sf{\longmapsto 774= 9a}$

$\sf{\longmapsto \cancel\dfrac{774}{9} = a }$

$\bf{\longmapsto 86 = a}$

The arithmetic expression of sequence is:-

• a = 86
• a+d = 86+(-34/3) = 224/3
• a+2d = 86 +2(-34/3) = 190/3

❝ Hence, the arithmetic expression of sequence is ❞

$\large{\boxed{\bf{\star \: 86, \dfrac{224}{3}, \dfrac{190}{3}..... \infty \: \star}}}$

______________________

Answer 2

UR QUESTION:-

ᴛʜᴇ sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ 10 ᴛᴇʀᴍs ᴏғ ᴀɴ ᴀʀɪᴛʜᴍᴇᴛɪᴄ sᴇϙᴜᴇɴᴄᴇ ɪs 350 ᴀɴᴅ ᴛʜᴇ sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ 15 ᴛᴇʀᴍs ɪs 100.ᴡʀɪᴛᴇ ᴛʜᴇ ᴀʟɢᴇʙʀᴀɪᴄ ᴇxᴘʀᴇssɪᴏɴ ғᴏʀ ᴛʜᴇ sᴇϙᴜᴇɴᴄᴇ?

UR ANSWER✓

$\Large\bold\purple{given,}$

$\sf\dashrightarrow sum\:of\:first\:10\:terms\:of\:A.P\:=350$

and,

$\sf\dashrightarrow sum\:for\:the\:first\:15\:terms=100$

$\Large\bold\red{TO\:FIND,}$

$\sf\dashrightarrow arithmetic\:expression\:seq$

$\Large\underline\bold{SOLUTION,}$

$\Large\underline\bold{ATQ........i.e.,.... according\:to\:the\:queston}$

$\sf\therefore taking\:two\:cases\:to\: understand\:better$

$\sf\implies in\:case1\:we\:will\:find\:sum\:of\:first\:10\:terms\:of\:A.P\:=350$

$\sf\implies in\:case2\:we\:will\:find\:the\:sum\:for\:the\:first\:15\:terms=100$

$\large {\fbox {CASE:-1}}$ $\sf\implies for\:seq\:350$

$\sf\therefore sum\:of\:first\:10\:terms\:of\:A.P\:=350........given$

$\sf\underline\bold{now,}$

$\sf\implies sum\:of\:the\:first\:n\:terms\:is\:given\:by,$

$\sf\large\therefore {S}_{n} = \dfrac{n}{2} [2a +(n-1)d]$

$\sf{\therefore {S}_{10} = 350}$

$\sf{\implies 350= \cancel\dfrac{10}{2}[2a +(10-1) d] }$

$\sf{\implies 350 = 5[2a + 9d]}$

$\sf{\implies \cancel\dfrac{350}{5} = 2a + 9d }$

$\bf{\implies 70 = 2a + 9d.........eq^1}$

$\sf{\implies a= \dfrac{70-9d}{2}}........eq^1$

$\large {\fbox {CASE:-2}}$ $\sf\implies for\:sum\:of\:15\:terms$

$\sf\therefore sum\:for\:the\:first\:15\:terms=100$

$\sf{\implies {S}_{15} = 100}$

$\sf{\implies 100= \cancel\dfrac{15}{2}[2a +(15-1) d] }$

$\sf{\implies 100 \times 2 = 15[2a + 14d]}$

$\sf{\implies \cancel\dfrac{200}{15} = 2a + 14d }$

$\sf{\implies \dfrac{40}{3} = 2a + 14d}$

$\sf{\implies 40 = 3(2a + 14d) }$

$\sf{\implies 40 = 6a + 42d }$

$\sf\therefore dividing\:the\:above\:eq\:by\:2$

we get.

$\sf{\implies 20 = 3a + 21d........eq^2 }$

$\sf\therefore substituting\:the\:value\:of\:eq^1\:in\:eq^2$

$\sf{\implies 20 = 3 \bigg(\dfrac{70-9d}{2} \bigg) + 21d}$

$\sf{\implies 20 = \dfrac{210 - 27d}{2} +21d }$

$\sf{\implies 20 = \dfrac{210-27d+42d}{2} }$

$\sf{\implies 40 = 210 + 15d}$

$\sf{\implies 40-210 = 15d }$

$\sf{\implies -170 = 15d }$

$\sf{\implies \cancel\dfrac{-170}{15} = d }$

$\sf{\implies d= \dfrac{-34}{3}........eq^3}$

$\sf\therefore substituting\:the\:value\:of\:eq^3\:in\:eq^2$

$\sf{\implies 20 = 3a + 21\bigg( \dfrac{-34}{3} \bigg)}$

$\sf{\implies 20 = 3a - \dfrac{-714}{3}}$

$\sf{\implies 20 = \dfrac{9a-714}{3}}$

$\sf{\implies 60 = 9a -714}$

$\sf{\implies 60+714 = 9a}$

$\sf{\implies 774= 9a}$

$\sf{\implies \cancel\dfrac{774}{9} = a }$

$\sf{\implies 86 = a}$

$\sf {\fbox { a=86}}$

the arithmetic expression of sequence is,

$\sf\therefore a=86$

$\sf\therefore a+d=86 + \bigg[ \dfrac{-34}{3} \bigg]= \dfrac{224}{3}$

a+2d = 86 +2(-34/3) = 190/3

$\sf\therefore a+2d= 86+ 2 \times \bigg( \dfrac{-34}{3} \bigg)= \dfrac{190}{3}$

Hence,

the arithmetic seq is,

$\sf{\boxed{\sf{86,\dfrac{224}{3}, \dfrac{190}{3}..... \:infinitily\:many}}}$