Question

please solve question 19 step by step ​

Answer 1

||✪✪ QUESTION ✪✪||

How many terms of series 3 + 7 + 11 ____ are Required to get a sum of 1176 ?

|| ✰✰ ANSWER ✰✰ ||

Here, we can see That,

→ First term = a = 3

→ common Difference = d = 7 - 3 = 4 .

→ Total terms = Let n .

→ Total sum = 1176 .

we know That :-

→ sum of n terms of AP = (n/2) [ 2a + (n-1)d ]

Putting all values we get :-

→ 1176 = (n/2) [ 2*3 + (n-1)4 ]

→ 2352 = n [ 6 + 4n - 4 ]

→ 2352 = 4n² + 2n

→ 4n² + 2n - 2352 = 0

→ 2(2n² + n - 1176) = 0

→ 2n² + n - 1176 = 0

Splitting The middle Term now,

→ 2n² - 48n + 49 n - 1176 = 0

→ 2n(n - 24) + 49( n - 24) = 0

→ (2n + 49) ( n - 24) = 0

Putting both Equal to Zero now,

→ (2n + 49) = 0

→ n = (-49)/2

Or,

→ ( n - 24) = 0

→ n = 24

Since, Negative value not possible.

Hence, Required Number of Terms Required To get a sum of 1176 of given Series is 24.

Answer 2

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$\huge\tt{TO~FIND:}$

• How many Terms of series 3 + 7 + 11 ..... are Required to get a sum of 1176 ?

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$\huge\tt{SOLUTION:}$

↪First term = a = 3

↪common Difference = D = 7 - 3 = 4

↪Total terms = n

↪Sum = 1176

We know that,

➜ sum of n terms of AP = (n/2) [ 2a + (n-1)d ]

If we put the values here,

↪1176 = (n/2) [ 2 × 3 + (n-1) 4 ]

↪2352 = n [ 6 + 4n - 4 ]

↪2352 = 4n² + 2n

↪4n² + 2n - 2352 = 0

↪2(2n² + n - 1176) = 0

↪2n² + n - 1176 = 0

↪2n² - 48n + 49 n - 1176 = 0

↪2n(n - 24) + 49( n - 24) = 0

↪(2n + 49) ( n - 24) = 0

↪(2n + 49) = 0

↪n = (-49)÷2

↪( n - 24) = 0

↪ n = 24

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${\huge{\sf{\fbox{\fbox{\red{N~=~24}}}}}}$

So, Number of Terms Required To get a sum of 1176 of given Series is 24

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