Question

In an A.P., if a = 6, d = 3, then the sum of first 10 terms is :- *

195

315

120

87

) For a quadratic equation, if the value of discriminant is 144, a = 1, b = -14, then its roots are :- (use formula method) *

13 and -1 are the roots of the equation

-13 and 1 are the roots of the equation

-13 and -1 are the roots of the equation

13 and 1 are the roots of the equation

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Answer 1

Answer:

• answer will be 315........................

Answer 2

Sum of first 10 terms of A.P is :- 165 .

Roots of quadratic equation are :- 13 and 1 .

Step-by-step explanation:

1st question :-

In an A.P., if a = 6, d = 3, then the sum of first 10 terms is :- 

Given :-

• a = 6
• d = 3
• n = 10

Where,

• a = first term
• d = common difference
• n = number of terms

To find :-

• The sum of 10 terms of A.P.

Formula used :-

★ S_n = n/2 [ 2a + ( n – 1 )d ]

Solution :-

S_n = n/2 [ 2a + ( n – 1 )d ]

Put the values of n , a and d in the formula we get :-

=> S_10 = 10/2 [ 2×6 + ( 10 – 1 )3 ]

=> S_10 = 5 ( 12 + 27 )

=> S_10 = 5 × 39

=> S_10 = 195 .

Hence , Sum of 10 terms of the A.P is 195.

option (a) is correct ✓✓

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2nd Question :- For a quadratic equation, if the value of discriminant is 144, a = 1, b = -14, then its roots are :- (use formula method)

Given :-

• Discriminant of quadratic equation = 144
• a = 1
• b = –14

To find :-

• The roots of the quadratic equation .

Formula used :-

★ roots of quadratic equation =$\underline{ \boxed{ \sf \: \dfrac{ - (b)± \sqrt{d} }{2a} } }$

Solution :-

Put the values of a,b and discriminant d in the formula, we get ;

roots of quadratic equation =$\sf \dfrac{ - ( - 14)± \sqrt{144} }{2 \times 1}$

$\sf : \implies \: \dfrac{14 \: ± \: 12}{2}$

$\sf : \implies \: \dfrac{14 - 12}{2} \: \: \: and\: \: \: \dfrac{14 + 12}{2}$

$\sf : \implies \: \dfrac{2}{2} \: \: \: and \: \: \: \dfrac{26}{2}$

$\sf : \implies \:1 \: and \: 13$

So, the roots are 1 and 13

option (d) is correct ✓✓